int(x^(3)+6x+1)/((x+2)^(2)(x^(2)+3))dx=

Partial Fraction Decomposition: To solve the integral of the given function, we will use partial fraction decomposition. The first step is to express the integrand as a sum of simpler fractions.Equation Expansion and Collection: The denominator (x + 2)^2 * (x^2 + 3) suggests that our partial fraction decomposition will have the form: A/(x + 2) + B/(x + 2)^2 + (Cx + D)/(x^2 + 3) where A, B, C, and D are constants that we need to determine.Coefficient Equations and Solution: We multiply both sides of the equation by the denominator to clear the fractions: x^3 + 6x + 1 = A(x + 2)(x^2 + 3) + B(x^2 + 3) + (Cx + D)(x + 2)^2Substitution and Simplification: Now we need to expand the right side and collect like terms to solve for A, B, C, and D.Final Values of Constants: Expanding the right side, we get: x^3 + 6x + 1 = Ax^3 + 3Ax + 2Ax^2 + 6A + Bx^2 + 3B + Cx^3 + 2Cx^2 + 4Cx + D + 4Dx + 4DIntegral Splitting: Combining like terms, we have: x^3 + 6x + 1 = (A + C)x^3 + (2A + B + 2C)x^2 + (3A + 4C + 4D)x + (6A + 3B + D)Logarithmic Integral: Now we equate the coefficients of the corresponding powers of x on both sides of the equation to solve for A, B, C, and D: 1. For x^3: A + C = 1 2. For x^2: 2A + B + 2C = 0 3. For x: 3A + 4C + 4D = 6 4. Constant term: 6A + 3B + D = 1Power Rule Integral: We have a system of four equations with four unknowns. We can solve this system using substitution or elimination methods.Substitution for Arctangent Integral: From equation 1, we get C = 1 - A.Combining All Parts: Substituting C = 1 - A into equation 2, we get: 2A + B + 2(1 - A) = 0 Simplifying, we find B = -2.Substituting C = 1 - A into equation 3, we get: 3A + 4(1 - A) + 4D = 6 Simplifying, we find 4D = 6 - 3A - 4 + 4A So, D = (2 + A)/4Substituting A = 1 - C and B = -2 into equation 4, we get: 6(1 - C) - 6 + D = 1 Simplifying, we find D = 1 + 6C - 6Now we have two expressions for D: D = (2 + A)/4 D = 1 + 6C - 6 Since C = 1 - A, we can substitute and solve for A.Substituting C = 1 - A into D = 1 + 6C - 6, we get: D = 1 + 6(1 - A) - 6 Simplifying, we find D = 1 - 6ANow we have two expressions for D: D = (2 + A)/4 D = 1 - 6A Equating them, we get: (2 + A)/4 = 1 - 6A Multiplying both sides by 4 to clear the fraction, we get: 2 + A = 4 - 24A Solving for A, we find A = 1/25Substituting A = 1/25 into C = 1 - A, we get C = 24/25.Substituting A = 1/25 into D = 1 - 6A, we get D = 1 - 6/25 = 19/25.Now we have the values of A, B, C, and D: A = 1/25, B = -2, C = 24/25, D = 19/25We can now write the partial fraction decomposition as: (1/25)/(x + 2) - 2/(x + 2)^2 + (24/25)x + 19/25)/(x^2 + 3)The integral can now be split into three parts: int(1/25)/(x + 2)dx - int2/(x + 2)^2dx + int((24/25)x + 19/25)/(x^2 + 3)dxThe first integral is a simple logarithmic integral: int(1/25)/(x + 2)dx = (1/25)ln|x + 2| + C1The second integral is a simple power rule integral: int2/(x + 2)^2dx = -2/(x + 2) + C2The third integral can be split into two parts: int(24/25)x/(x^2 + 3)dx + int(19/25)/(x^2 + 3)dxFor the first part of the third integral, we use a substitution: Let u = x^2 + 3, then du = 2xdx.Substituting, we get: int(24/25)x/(x^2 + 3)dx = (24/50)int1/udu = (24/50)ln|u| + C3 Substituting back, we get: (24/50)ln|x^2 + 3| + C3For the second part of the third integral, we have an arctangent integral: int(19/25)/(x^2 + 3)dx = (19/25)(1/sqrt(3))arctan(x/sqrt(3)) + C4Combining all the parts, we get the final answer: (1/25)ln|x + 2| - 2/(x + 2) + (24/50)ln|x^2 + 3| + (19/25)(1/sqrt(3))arctan(x/sqrt(3)) + C where C = C1 + C2 + C3 + C4 is the constant of integration.

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\int \frac{x^{3}+6 x+1}{(x+2)^{2}\left(x^{2}+3\right)} d x

Partial Fraction Decomposition: To solve the integral of the given function, we will use partial fractions" target="_blank" class="backlink">fraction decomposition. The first step is to express the integrand as a sum of simpler fractions.
Equation Expansion and Collection: The denominator $(x + 2)^2 * (x^2 + 3)$ suggests that our partial fraction decomposition will have the form: $\newline$ $\frac{A}{(x + 2)} + \frac{B}{(x + 2)^2} + \frac{Cx + D}{(x^2 + 3)}$ $\newline$ where $A$ , $B$ , $C$ , and $D$ are constants that we need to determine.
Coefficient Equations and Solution: We multiply both sides of the equation by the denominator to clear the fractions: $x^3 + 6x + 1 = A(x + 2)(x^2 + 3) + B(x^2 + 3) + (Cx + D)(x + 2)^2$
Substitution and Simplification: Now we need to expand the right side and collect like terms to solve for $A$ , $B$ , $C$ , and $D$ .
Final Values of Constants: Expanding the right side, we get: $\newline$ $x^3 + 6x + 1 = Ax^3 + 3Ax + 2Ax^2 + 6A + Bx^2 + 3B + Cx^3 + 2Cx^2 + 4Cx + D + 4Dx + 4D$
Integral Splitting: Combining like terms, we have: $x^3 + 6x + 1 = (A + C)x^3 + (2A + B + 2C)x^2 + (3A + 4C + 4D)x + (6A + 3B + D)$
Logarithmic Integral: Now we equate the coefficients of the corresponding powers of $x$ on both sides of the equation to solve for $A$ , $B$ , $C$ , and $D$ :
$1$ . For $x^3$ : $A + C = 1$
$2$ . For $x^2$ : $2A + B + 2C = 0$
$3$ . For $x$ : $A$ $0$
$4$ . Constant term: $A$ $1$
Power Rule Integral: We have a system of four equations with four unknowns. We can solve this system using substitution or elimination methods.
Substitution for Arctangent Integral: From equation $1$ , we get $C = 1 - A$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $0$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $2A + B + 2(1 - A) = 0$ $5$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $0$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $2A + B + 2(1 - A) = 0$ $5$ .Substituting $C = 1 - A$ into $2A + B + 2(1 - A) = 0$ $0$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $8$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $9$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $0$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $2A + B + 2(1 - A) = 0$ $5$ .Substituting $C = 1 - A$ into $2A + B + 2(1 - A) = 0$ $0$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $8$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $9$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $9$ $\newline$ Equating them, we get: $\newline$ $B = -2$ $3$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $B = -2$ $4$ $\newline$ Solving for $2A + B + 2(1 - A) = 0$ $5$ , we find $B = -2$ $6$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $0$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $2A + B + 2(1 - A) = 0$ $5$ .Substituting $C = 1 - A$ into $2A + B + 2(1 - A) = 0$ $0$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $8$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $9$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $9$ $\newline$ Equating them, we get: $\newline$ $B = -2$ $3$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $B = -2$ $4$ $\newline$ Solving for $2A + B + 2(1 - A) = 0$ $5$ , we find $B = -2$ $6$ Substituting $B = -2$ $6$ into $C = 1 - A$ , we get $B = -2$ $9$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $0$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $0$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $2A + B + 2(1 - A) = 0$ $5$ .Substituting $C = 1 - A$ into $2A + B + 2(1 - A) = 0$ $0$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $8$ $\newline$ Simplifying, we find $2A + B + 2(1 - A) = 0$ $9$ Now we have two expressions for $2A + B + 2(1 - A) = 0$ $1$ : $\newline$ $D = (2 + A)/4$ $\newline$ $2A + B + 2(1 - A) = 0$ $9$ $\newline$ Equating them, we get: $\newline$ $B = -2$ $3$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $B = -2$ $4$ $\newline$ Solving for $2A + B + 2(1 - A) = 0$ $5$ , we find $B = -2$ $6$ Substituting $B = -2$ $6$ into $C = 1 - A$ , we get $B = -2$ $9$ .Substituting $B = -2$ $6$ into $2A + B + 2(1 - A) = 0$ $9$ , we get $C = 1 - A$ $2$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $8$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $8$ The third integral can be split into two parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $9$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = \frac{2 + A}{4}$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = \frac{2 + A}{4}$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = \frac{2 + A}{4}$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $\frac{2 + A}{4} = 1 - 6A$ $\newline$ Multiplying both sides by $B = -2$ $9$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $1$ Substituting $C = 1 - A$ $1$ into $C = 1 - A$ , we get $C = 1 - A$ $4$ .Substituting $C = 1 - A$ $1$ into $B = -2$ $5$ , we get $C = 1 - A$ $7$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $9$ , $4D = 6 - 3A - 4 + 4A$ $0$ , and $D$ : $\newline$ $C = 1 - A$ $1$ , $B = -2$ , $C = 1 - A$ $4$ , $4D = 6 - 3A - 4 + 4A$ $5$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $8$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $9$ The third integral can be split into two parts: $\newline$ $D = \frac{2 + A}{4}$ $0$ For the first part of the third integral, we use a substitution: $\newline$ Let $D = \frac{2 + A}{4}$ $1$ , then $D = \frac{2 + A}{4}$ $2$ .
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $8$ The third integral can be split into two parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $9$ For the first part of the third integral, we use a substitution: $\newline$ Let $D = (2 + A)/4$ $0$ , then $D = (2 + A)/4$ $1$ .Substituting, we get: $\newline$ $D = (2 + A)/4$ $2$ $\newline$ Substituting back, we get: $\newline$ $D = (2 + A)/4$ $3$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = (2 + A)/4$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = (2 + A)/4$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $(2 + A)/4 = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ The first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $8$ The third integral can be split into two parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $9$ For the first part of the third integral, we use a substitution: $\newline$ Let $D = (2 + A)/4$ $0$ , then $D = (2 + A)/4$ $1$ .Substituting, we get: $\newline$ $D = (2 + A)/4$ $2$ $\newline$ Substituting back, we get: $\newline$ $D = (2 + A)/4$ $3$ For the second part of the third integral, we have an arctangent integral: $\newline$ $D = (2 + A)/4$ $4$
Combining All Parts: Substituting $C = 1 - A$ into equation $2$ , we get: $\newline$ $2A + B + 2(1 - A) = 0$ $\newline$ Simplifying, we find $B = -2$ .Substituting $C = 1 - A$ into equation $3$ , we get: $\newline$ $3A + 4(1 - A) + 4D = 6$ $\newline$ Simplifying, we find $4D = 6 - 3A - 4 + 4A$ $\newline$ So, $D = \frac{2 + A}{4}$ Substituting $A = 1 - C$ and $B = -2$ into equation $4$ , we get: $\newline$ $6(1 - C) - 6 + D = 1$ $\newline$ Simplifying, we find $D = 1 + 6C - 6$ Now we have two expressions for $D$ : $\newline$ $D = \frac{2 + A}{4}$ $\newline$ $D = 1 + 6C - 6$ $\newline$ Since $C = 1 - A$ , we can substitute and solve for $B = -2$ $2$ .Substituting $C = 1 - A$ into $D = 1 + 6C - 6$ , we get: $\newline$ $D = 1 + 6(1 - A) - 6$ $\newline$ Simplifying, we find $B = -2$ $5$ Now we have two expressions for $D$ : $\newline$ $D = \frac{2 + A}{4}$ $\newline$ $B = -2$ $5$ $\newline$ Equating them, we get: $\newline$ $\frac{2 + A}{4} = 1 - 6A$ $\newline$ Multiplying both sides by $4$ to clear the fraction, we get: $\newline$ $2 + A = 4 - 24A$ $\newline$ Solving for $B = -2$ $2$ , we find $C = 1 - A$ $0$ Substituting $C = 1 - A$ $0$ into $C = 1 - A$ , we get $C = 1 - A$ $3$ .Substituting $C = 1 - A$ $0$ into $B = -2$ $5$ , we get $C = 1 - A$ $6$ .Now we have the values of $B = -2$ $2$ , $C = 1 - A$ $8$ , $C = 1 - A$ $9$ , and $D$ : $\newline$ $C = 1 - A$ $0$ , $B = -2$ , $C = 1 - A$ $3$ , $4D = 6 - 3A - 4 + 4A$ $4$ We can now write the partial fraction decomposition as: $\newline$ $4D = 6 - 3A - 4 + 4A$ $5$ /(x + $2$ ) - \frac{ $2$ }{(x + $2$ )^ $2$ } + \frac{( $24$ / $25$ )x + $19$ / $25$ }{(x^ $2$ + $3$ )}\)The integral can now be split into three parts: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ /(x + $2$ )dx - $4D = 6 - 3A - 4 + 4A$ $7$ /(x + $2$ )^ $2$ dx + $4D = 6 - 3A - 4 + 4A$ $8$ dxThe first integral is a simple logarithmic integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $6$ /(x + $2$ )dx = \frac{ $1$ }{ $25$ }\ln|x + $2$ | + C $1$ The second integral is a simple power rule integral: $\newline$ $4D = 6 - 3A - 4 + 4A$ $7$ /(x + $2$ )^ $2$ dx = -\frac{ $2$ }{(x + $2$ )} + C $2$ The third integral can be split into two parts: $\newline$ $D = \frac{2 + A}{4}$ $1$ /(x^ $2$ + $3$ )dx + $D = \frac{2 + A}{4}$ $2$ /(x^ $2$ + $3$ )dxFor the first part of the third integral, we use a substitution: $\newline$ Let $D = \frac{2 + A}{4}$ $3$ , then $D = \frac{2 + A}{4}$ $4$ .Substituting, we get: $\newline$ $D = \frac{2 + A}{4}$ $1$ /(x^ $2$ + $3$ )dx = \frac{ $24$ }{ $50$ }\int \frac{ $1$ }{u}du = \frac{ $24$ }{ $50$ }\ln|u| + C $3$ $\newline$ Substituting back, we get: $\newline$ $D = \frac{2 + A}{4}$ $6$ For the second part of the third integral, we have an arctangent integral: $\newline$ $D = \frac{2 + A}{4}$ $2$ /(x^ $2$ + $3$ )dx = \frac{ $19$ }{ $25$ }(\frac{ $1$ }{\sqrt{ $3$ }})\arctan(\frac{x}{\sqrt{ $3$ }}) + C $4$ Combining all the parts, we get the final answer: $\newline$ $D = \frac{2 + A}{4}$ $8$ $\newline$ where $D = \frac{2 + A}{4}$ $9$ is the constant of integration.

$\int \frac{x^{3}+6 x+1}{(x+2)^{2}\left(x^{2}+3\right)} d x$ =

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∫x3+6x+1(x+2)2(x2+3)dx \int \frac{x^{3}+6 x+1}{(x+2)^{2}\left(x^{2}+3\right)} d x ∫(x+2)2(x2+3)x3+6x+1​dx=

$\int \frac{x^{3}+6 x+1}{(x+2)^{2}\left(x^{2}+3\right)} d x$ =