Evaluate the integral int(x+4)/(2x+6)dx. Choose 1 answer: (A) 2x-(ln |2x+6|)/(2)+C (B) (1)/(2)x-(ln |2x+6|)/(2)+C (C) 2x+ln |2x+6|+C (D) (1)/(2)x+(ln |2x+6|)/(2)+C

Partial Fraction Decomposition: Let's do a partial fraction decomposition first. We can write (x+4)/(2x+6) as A/(2x+6) + B, where A and B are constants we need to find.Finding Constants A and B: To find A and B, we set x+4 = A(1) + B(2x+6). Let's choose x = -3 to make the B term disappear and solve for A.Solving for A: Plugging x = -3 into the equation, we get -3+4 = A(1) + B(0), which simplifies to A = 1.Finding B: Now let's find B. We can choose x = 0 to make the A term disappear. Plugging x = 0 into the equation, we get 0+4 = A(0) + B(2*0+6), which simplifies to B = 2/3.Final Integration: We now have the partial fraction decomposition: (x+4)/(2x+6) = 1/(2x+6) + (2/3). Let's integrate both terms separately.Integral of 1/(2x+6): The integral of 1/(2x+6) is (1/2)ln|2x+6|, because the derivative of 2x+6 is 2, and we need to multiply by 1/2 to compensate for the extra 2.Integral of 2/3: The integral of 2/3 is (2/3)x, because the integral of a constant is just the constant times x.Combining Integrals: Adding both integrals together, we get (1/2)ln|2x+6| + (2/3)x + C.

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Evaluate the integral
int(x+4)/(2x+6)dx.
Choose 1 answer:
(A)
2x-(ln |2x+6|)/(2)+C
(B)
(1)/(2)x-(ln |2x+6|)/(2)+C
(C)
2x+ln |2x+6|+C
(D)
(1)/(2)x+(ln |2x+6|)/(2)+C

Evaluate the integral $\int \frac{x+4}{2 x+6} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $2 x-\frac{\ln |2 x+6|}{2}+C$ $\newline$ (B) $\frac{1}{2} x-\frac{\ln |2 x+6|}{2}+C$ $\newline$ (C) $2 x+\ln |2 x+6|+C$ $\newline$ (D) $\frac{1}{2} x+\frac{\ln |2 x+6|}{2}+C$

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Calculus

Evaluate definite integrals using the chain rule

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Q. Evaluate the integral

\int \frac{x+4}{2 x+6} d x

\newline

Choose

1

answer:

\newline

(A)

2 x-\frac{\ln |2 x+6|}{2}+C

\newline

(B)

\frac{1}{2} x-\frac{\ln |2 x+6|}{2}+C

\newline

(C)

2 x+\ln |2 x+6|+C

\newline

(D)

\frac{1}{2} x+\frac{\ln |2 x+6|}{2}+C

Partial Fraction Decomposition: Let's do a partial fraction decomposition first. We can write $(x+4)/(2x+6)$ as $A/(2x+6) + B$ , where $A$ and $B$ are constants we need to find.
Finding Constants $A$ and $B$ : To find $A$ and $B$ , we set $x+4 = A(1) + B(2x+6)$ . Let's choose $x = -3$ to make the $B$ term disappear and solve for $A$ .
Solving for $A$ : Plugging $x = -3$ into the equation, we get $-3+4 = A(1) + B(0)$ , which simplifies to $A = 1$ .
Finding B: Now let's find $B$ . We can choose $x = 0$ to make the $A$ term disappear. Plugging $x = 0$ into the equation, we get $0+4 = A(0) + B(2\times0+6)$ , which simplifies to $B = \frac{2}{3}$ .
Final Integration: We now have the partial fraction decomposition: $(x+4)/(2x+6) = 1/(2x+6) + (2/3)$ . Let's integrate both terms separately.
Integral of $\frac{1}{2x+6}$ : The integral of $\frac{1}{2x+6}$ is $(\frac{1}{2})\ln|2x+6|$ , because the derivative of $2x+6$ is $2$ , and we need to multiply by $\frac{1}{2}$ to compensate for the extra $2$ .
Integral of $\frac{2}{3}$ : The integral of $\frac{2}{3}$ is $(\frac{2}{3})x$ , because the integral of a constant is just the constant times $x$ .
Combining Integrals: Adding both integrals together, we get $(\frac{1}{2})\ln|2x+6| + (\frac{2}{3})x + C$ .