Question

Evaluate the following integrals:

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\log x}{{\left(x+1\right)}^2}\mathrm{d}x \\ \mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{be}\left(\log x\right)\mathrm{and}\mathrm{second}\mathrm{function}\mathrm{be}\frac{1}{{\left(x+1\right)}^2}. \\ \mathrm{First}\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{integral}\mathrm{of}\mathrm{the}\mathrm{second}\mathrm{function},\mathrm{i}.\mathrm{e}.,\int \frac{1}{{\left(x+1\right)}^2}\mathrm{d}x. \\ \mathrm{Put}t=\left(x+1\right).\mathrm{Then}\mathrm{d}t=\mathrm{d}x \\ \mathrm{Therefore}, \\ \int \frac{1}{{\left(x+1\right)}^2}\mathrm{d}x=\int t^{-2}\mathrm{d}t \\ =-\frac{1}{t} \\ =-\frac{1}{1+x} \\ \mathrm{Hence},\mathrm{using}\mathrm{integration}\mathrm{by}\mathrm{parts},\mathrm{we}\mathrm{get} \\ \int \frac{\log x}{{\left(x+1\right)}^2}\mathrm{d}x=\left(\log x\right)\int \frac{1}{{\left(x+1\right)}^2}\mathrm{d}x-\int \left[\left(\frac{d\left(\log x\right)}{dx}\right)\int \frac{1}{{\left(x+1\right)}^2}\mathrm{d}x\right]\mathrm{d}x \\ =\left(\log x\right)\left(-\frac{1}{1+x}\right)-\int \left(\frac{1}{x}\right)\left(-\frac{1}{1+x}\right)dx \\ =-\frac{\log x}{1+x}+\int \left(\frac{1}{x^2+x}\right)dx \\ =-\frac{\log x}{1+x}+\int \frac{1}{x^2+x+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}}dx \\ =-\frac{\log x}{1+x}+\int \frac{1}{{\left(x+{\displaystyle \frac{1}{2}}\right)}^2-{\displaystyle {\left(\frac{1}{2}\right)}^2}}dx \\ =-\frac{\log x}{1+x}+\frac{1}{2\times {\displaystyle \frac{1}{2}}}\log \left|\frac{x+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{2}}}{x+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}}\right|+c \\ =-\frac{\log x}{1+x}+\log \left|\frac{x}{x+1}\right|+c \\ \mathrm{Hence},\int \frac{\log x}{{\left(x+1\right)}^2}\mathrm{d}x=-\frac{\log x}{1+x}+\log \left|\frac{x}{x+1}\right|+c \end{gathered}$$