Question

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

Answer 1

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

$\log x=t,\Rightarrow x=e^t$

$\Rightarrow \frac{1}{x}dx=dt$

$\Rightarrow dx=xdt$

$\therefore I=\int \frac{t}{{\left(1+t\right)}^2}xdt=\int \frac{e^t.t}{{\left(1+t\right)}^2}dt$

$I=\int \frac{e^t.(t+1-1)}{{\left(1+t\right)}^2}dt$

$I=\int \frac{e^t.(t+1)}{{\left(1+t\right)}^2}dt-\int \frac{e^t}{{\left(1+t\right)}^2}dt$

$I=\int e^t[\frac{1}{1+t}-\frac{1}{1+t^2}]dt$

$\left\{e^x\right[f\left(x\right)+f^{\prime }\left(x\right)]=e^{x}f\left(x\right)+C\}$

$I=e^t\frac{1}{1+t}+C$

$I=\frac{x}{1+\log x}+C$