The value of int {frac{{log x}}{{{{(x + 1)}^2 | vedclass

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(A) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \log x$ and $dv = (x + 1)^{-2} dx$. Then

Vedclass · Accepted Answer

$\frac{{ - \log x}}{{x + 1}} + \log x - \log \,(x + 1)$

Vedclass · Answer

(A) We use the method of integration by parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$. Let $u = \log x$ and $dv = (x + 1)^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -(x + 1)^{-1}$. Applying the formula: $\int \frac{\log x}{(x + 1)^2} dx = \log x \cdot (-(x + 1)^{-1}) - \int \frac{1}{x} \cdot (-(x + 1)^{-1}) dx$ $= -\frac{\log x}{x + 1} + \int \frac{1}{x(x + 1)} dx$. Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$. So,$\int \frac{1}{x(x + 1)} dx = \int \left( \frac{1}{x} - \frac{1}{x + 1} \right) dx = \log |x| - \log |x + 1| + C$. Therefore,the integral is $-\frac{\log x}{x + 1} + \log x - \log (x + 1) + C$.

The value of $\int {\frac{{\log x}}{{{{(x + 1)}^2}}}dx} $ is

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