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∫1x3+1 dx=\int \frac{1}{x^{3}+1} \, dx =

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Evaluate definite integrals using the power ruleright chevron icon

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Q. ∫1x3+1 dx=\int \frac{1}{x^{3}+1} \, dx =
  1. Recognize Problem Complexity: Recognize that the integral of 1x3+1\frac{1}{x^3 + 1} is not a standard integral and will likely require partial fraction decomposition.
  2. Factor Denominator: Factor the denominator x3+1x^3 + 1. This can be factored using the sum of cubes formula: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2), where a=xa = x and b=1b = 1.\newlinex3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1)(x^2 - x + 1)
  3. Set Up Partial Fractions: Set up the partial fraction decomposition.\newline1x3+1=1(x+1)(x2−x+1)=Ax+1+Bx+Cx2−x+1\frac{1}{x^3 + 1} = \frac{1}{(x + 1)(x^2 - x + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1}
  4. Clear Fractions: Multiply both sides by the common denominator (x+1)(x2−x+1)(x + 1)(x^2 - x + 1) to clear the fractions.\newline1=A(x2−x+1)+(Bx+C)(x+1)1 = A(x^2 - x + 1) + (Bx + C)(x + 1)
  5. Expand and Collect Terms: Expand the right side and collect like terms.\newline1=Ax2−Ax+A+Bx2+Bx+Cx+C1 = Ax^2 - Ax + A + Bx^2 + Bx + Cx + C\newline1=(A+B)x2+(−A+B+C)x+(A+C)1 = (A + B)x^2 + (-A + B + C)x + (A + C)
  6. Equate Coefficients: Equate the coefficients of corresponding powers of xx on both sides of the equation.\newlineFor x2x^2: A+B=0A + B = 0\newlineFor xx: −A+B+C=0-A + B + C = 0\newlineFor the constant term: A+C=1A + C = 1
  7. Solve System of Equations: Solve the system of equations for AA, BB, and CC. From A+B=0A + B = 0, we get B=−AB = -A. Substitute B=−AB = -A into −A+B+C=0-A + B + C = 0 to get −A−A+C=0-A - A + C = 0, which simplifies to −2A+C=0-2A + C = 0. Now we have two equations: A+C=1A + C = 1 −2A+C=0-2A + C = 0
  8. Find Values of A, B, C: Solve the system of equations.\newlineSubtract the second equation from the first to eliminate C:\newlineA+C−(−2A+C)=1−0A + C - (-2A + C) = 1 - 0\newlineA+C+2A−C=1A + C + 2A - C = 1\newline3A=13A = 1\newlineA=13A = \frac{1}{3}
  9. Write Partial Fraction Decomposition: Find BB and CC using the value of AA.\newlineSince B=−AB = -A, B=−13B = -\frac{1}{3}.\newlineSubstitute AA into −2A+C=0-2A + C = 0 to find CC:\newline−2(13)+C=0-2(\frac{1}{3}) + C = 0\newline−23+C=0-\frac{2}{3} + C = 0\newlineCC00
  10. Integrate Each Term: Write the partial fraction decomposition with the found values of AA, BB, and CC.1x3+1=13x+1+−13x+23x2−x+1\frac{1}{x^3 + 1} = \frac{\frac{1}{3}}{x + 1} + \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1}
  11. Integrate First Term: Integrate each term separately.\newline∫131x+1dx+∫−13x+23x2−x+1dx\int\frac{1}{3}\frac{1}{x + 1} dx + \int\frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} dx
  12. Complete the Square: The first integral is straightforward.\newline∫131x+1dx=13ln⁡∣x+1∣+C1\int\frac{1}{3}\frac{1}{x + 1} dx = \frac{1}{3}\ln|x + 1| + C_1
  13. Rewrite Integral: The second integral requires completing the square for the quadratic denominator. x2−x+1x^2 - x + 1 can be written as (x−12)2+34(x - \frac{1}{2})^2 + \frac{3}{4} by completing the square.
  14. Split Integral: Rewrite the second integral with the completed square. ∫−13x+23(x−12)2+(34)dx\int\frac{-\frac{1}{3}x + \frac{2}{3}}{(x - \frac{1}{2})^2 + (\frac{3}{4})} dx
  15. Integrate First Part: Split the second integral into two parts.\newline\int\frac{-\frac{\(1\)}{\(3\)}x}{(x - \frac{\(1\)}{\(2\)})^\(2\) + \frac{\(3\)}{\(4\)}} dx + \int\frac{\frac{\(2\)}{\(3\)}}{(x - \frac{\(1\)}{\(2\)})^\(2\) + \frac{\(3\)}{\(4\)}} dx
  16. Integrate Second Part: The first part of the second integral is a standard integral that results in an arctangent function after a substitution, and the second part is a logarithmic function.\(\newlineLet u=x−12u = x - \frac{1}{2}, then du=dxdu = dx, and the integral becomes:\newline−13∫u(u)2+(34)du+23∫1(u)2+(34)du-\frac{1}{3} \int\frac{u}{(u)^2 + (\frac{3}{4})} du + \frac{2}{3} \int\frac{1}{(u)^2 + (\frac{3}{4})} du
  17. Combine Integrals: Integrate the first part using the arctangent formula.\newline−13∫uu2+34du=−13⋅23⋅arctan(2u3)+C2-\frac{1}{3} \int \frac{u}{u^2 + \frac{3}{4}} du = -\frac{1}{3} \cdot \frac{2}{\sqrt{3}} \cdot \text{arctan}\left(\frac{2u}{\sqrt{3}}\right) + C_2
  18. Combine Constants: Integrate the second part using the logarithmic formula.\newline23∫1(u)2+(34)du=23×23×ln⁡∣u+32∣−ln⁡∣u−32∣+C3\frac{2}{3} \int \frac{1}{(u)^2 + (\frac{3}{4})} du = \frac{2}{3} \times \frac{2}{\sqrt{3}} \times \ln|u + \frac{\sqrt{3}}{2}| - \ln|u - \frac{\sqrt{3}}{2}| + C_3
  19. Combine Constants: Integrate the second part using the logarithmic formula.\newline23∫1(u)2+(34)du=23⋅23⋅ln⁡∣u+32∣−ln⁡∣u−32∣+C\frac{2}{3} \int \frac{1}{(u)^2 + (\frac{3}{4})} du = \frac{2}{3} \cdot \frac{2}{\sqrt{3}} \cdot \ln|u + \frac{\sqrt{3}}{2}| - \ln|u - \frac{\sqrt{3}}{2}| + CCombine the results of the integrals and substitute back for u.\newline∫131x+1dx+∫−13x+23x2−x+1dx=13ln⁡∣x+1∣−233arctan⁡(2(x−12)3)+433ln⁡∣(x−12)+32∣−ln⁡∣(x−12)−32∣+C\int \frac{1}{3}\frac{1}{x + 1} dx + \int \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} dx = \frac{1}{3}\ln|x + 1| - \frac{2}{3}\sqrt{3} \arctan\left(\frac{2(x - \frac{1}{2})}{\sqrt{3}}\right) + \frac{4}{3}\sqrt{3} \ln|\left(x - \frac{1}{2}\right) + \frac{\sqrt{3}}{2}| - \ln|\left(x - \frac{1}{2}\right) - \frac{\sqrt{3}}{2}| + C
  20. Combine Constants: Integrate the second part using the logarithmic formula.\newline23∫1(u)2+(34)du=23⋅23⋅ln⁡∣u+32∣−ln⁡∣u−32∣+C\frac{2}{3} \int \frac{1}{(u)^2 + (\frac{3}{4})} du = \frac{2}{3} \cdot \frac{2}{\sqrt{3}} \cdot \ln|u + \frac{\sqrt{3}}{2}| - \ln|u - \frac{\sqrt{3}}{2}| + C Combine the results of the integrals and substitute back for u.\newline∫131x+1dx+∫−13x+23x2−x+1dx=13ln⁡∣x+1∣−233arctan⁡(2(x−12)3)+433ln⁡∣(x−12)+32∣−ln⁡∣(x−12)−32∣+C\int \frac{1}{3}\frac{1}{x + 1} dx + \int \frac{-\frac{1}{3}x + \frac{2}{3}}{x^2 - x + 1} dx = \frac{1}{3}\ln|x + 1| - \frac{2}{3}\sqrt{3} \arctan\left(\frac{2(x - \frac{1}{2})}{\sqrt{3}}\right) + \frac{4}{3}\sqrt{3} \ln|\left(x - \frac{1}{2}\right) + \frac{\sqrt{3}}{2}| - \ln|\left(x - \frac{1}{2}\right) - \frac{\sqrt{3}}{2}| + C Combine the constants of integration into a single constant C.\newlineFinal answer: ∫1x3+1dx=13ln⁡∣x+1∣−233arctan⁡(2(x−12)3)+433ln⁡∣(x−12)+32∣−ln⁡∣(x−12)−32∣+C\int \frac{1}{x^3 + 1} dx = \frac{1}{3}\ln|x + 1| - \frac{2}{3}\sqrt{3} \arctan\left(\frac{2(x - \frac{1}{2})}{\sqrt{3}}\right) + \frac{4}{3}\sqrt{3} \ln|\left(x - \frac{1}{2}\right) + \frac{\sqrt{3}}{2}| - \ln|\left(x - \frac{1}{2}\right) - \frac{\sqrt{3}}{2}| + C

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