# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{\log x}{(x+1)^{2}} d x$

Solution:

We know that integration by parts is given by:

$\int \mathrm{UV}=\mathrm{U} \int \mathrm{V} \mathrm{dv}-\int \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{U} \int \mathrm{V} \mathrm{dv}$

Choosing $\log x$ as first function and $\frac{1}{(x+1)^{2}}$ as second function we get,

$\int \frac{\log x}{(x+1)^{2}} d x=\log x \int\left(\frac{1}{(x+1)^{2}}\right) d x-\int\left(\frac{d}{d x}(\log x) \int \frac{1}{(x+1)^{2}} d x\right) d x$

$\int \frac{\log x}{(x+1)^{2}} d x=\log x\left(-\frac{1}{x+1}\right)+\int \frac{1}{x}\left(\frac{1}{x+1}\right) d x$

$\int \frac{\log x}{(x+1)^{2}} d x=-\frac{\log x}{x+1}+\int \frac{(x+1)-(x)}{x(x+1)} d x$

$\int \frac{\log x}{(x+1)^{2}} d x=-\frac{\log x}{x+1}+\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x$

$\int \frac{\log x}{(x+1)^{2}} d x=-\frac{\log x}{x+1}+\log x-\log (x+1)+c$

$\int \frac{\log x}{(x+1)^{2}} d x=-\frac{\log x}{x+1}+\log \left(\frac{x}{x+1}\right)+c$