int frac{log x}{(1+x)^3} d x= | vedclass

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(D) Let $I = \int \frac{\log x}{(1+x)^3} dx$. Using integration by parts, let $u = \log x$ and $dv = (1+x)^{-3} dx$. Then $du = \frac{1}{x} dx$ and $v

Vedclass · Accepted Answer

$\frac{1}{2}\left[\frac{1}{1+x}-\frac{\log x}{(1+x)^2}+\log \left(\frac{x}{1+x}\right)\right]+c$

Vedclass · Answer

(D) Let $I = \int \frac{\log x}{(1+x)^3} dx$. Using integration by parts, let $u = \log x$ and $dv = (1+x)^{-3} dx$. Then $du = \frac{1}{x} dx$ and $v = \int (1+x)^{-3} dx = \frac{(1+x)^{-2}}{-2} = -\frac{1}{2(1+x)^2}$. Applying the formula $\int u dv = uv - \int v du$: $I = -\frac{\log x}{2(1+x)^2} - \int \left(-\frac{1}{2(1+x)^2}\right) \frac{1}{x} dx$ $I = -\frac{\log x}{2(1+x)^2} + \frac{1}{2} \int \frac{1}{x(1+x)^2} dx$. Using partial fractions for $\frac{1}{x(1+x)^2} = \frac{A}{x} + \frac{B}{1+x} + \frac{C}{(1+x)^2}$: $1 = A(1+x)^2 + Bx(1+x) + Cx$. For $x=0$, $A=1$. For $x=-1$, $C=-1$. Comparing coefficients of $x^2$: $A+B=0 \implies B=-1$. So, $\int \frac{1}{x(1+x)^2} dx = \int \left(\frac{1}{x} - \frac{1}{1+x} - \frac{1}{(1+x)^2}\right) dx = \log|x| - \log|1+x| + \frac{1}{1+x} = \log\left|\frac{x}{1+x}\right| + \frac{1}{1+x}$. Substituting back: $I = -\frac{\log x}{2(1+x)^2} + \frac{1}{2} \left(\log\left|\frac{x}{1+x}\right| + \frac{1}{1+x}\right) + c = \frac{1}{2} \left[ \frac{1}{1+x} - \frac{\log x}{(1+x)^2} + \log\left|\frac{x}{1+x}\right| \right] + c$.

$\int \frac{\log x}{(1+x)^3} d x=$

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