The value of the integral int frac{sin(ln(2 + | vedclass

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Question

(A) Let $I = \int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$. We can rewrite the argument of the logarithm as $2 + 2x = 2(1 + x)$. So,$I = \int \frac{\sin(\l

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$- \cos(\ln(2x + 2)) + C$

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(A) Let $I = \int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$. We can rewrite the argument of the logarithm as $2 + 2x = 2(1 + x)$. So,$I = \int \frac{\sin(\ln(2(1 + x)))}{x + 1} dx = \int \frac{\sin(\ln 2 + \ln(1 + x))}{x + 1} dx$. Let $u = \ln(2 + 2x)$. Then $du = \frac{1}{2 + 2x} \cdot 2 dx = \frac{1}{1 + x} dx$. Substituting these into the integral,we get $I = \int \sin(u) du$. The integral of $\sin(u)$ is $-\cos(u) + C$. Substituting back $u = \ln(2 + 2x)$,we get $I = -\cos(\ln(2 + 2x)) + C$.

The value of the integral $\int \frac{\sin(\ln(2 + 2x))}{x + 1} dx$ is

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