Question

Evaluate $\int lnx.\frac{1}{{\left(x+1\right)}^2}dx$

Answer 1

$\int \ln x\frac{1}{{(x+1)}^2}dx$

Let $u=\ln x\Rightarrow du=\frac{1}{x}dx$

$dv=\frac{1}{{(x+1)}^2}dx\Rightarrow v=-\frac{1}{x+1}$

$=-\frac{\ln x}{x+1}-\int -\frac{1}{x(x+1)}dx$

$=-\frac{\ln x}{x+1}+\int \frac{x+1-x}{x(x+1)}dx$

$=-\frac{\ln x}{x+1}+\int \frac{x+1}{x(x+1)}dx-\int \frac{x}{x(x+1)}dx$

$=-\frac{\ln x}{x+1}+\int \frac{dx}{x}-\int \frac{dx}{(x+1)}$

$=-\frac{\ln x}{x+1}+\ln \left|x\right|-\ln |x+1|+c$

$=-\frac{\ln x}{x+1}+\ln \left|\frac{x}{x+1}\right|+c$