Question

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

Answer 1

$\begin{array}{c}\int \frac{\mathrm{logx}}{{\left(1+\log x\right)}^2}dx\\ Let,\log x=t\\ x=e^t\\ dx=xdt\\ \therefore I=\int \frac{t}{{\left(1+t\right)}^2}xdt=\int \frac{e^t.t}{{\left(1+t\right)}^2}dt=\int \frac{e^t.\left(t+1-1\right)}{{\left(1+t\right)}^2}dt\\ =\int \frac{e^t.\left(t+1\right)}{{\left(1+t\right)}^2}dt-\int \frac{e^t}{\left(1+t^2\right)}dt\\ =\int e^t\left[\frac{1}{1+t}-\frac{1}{1+t^2}\right]dx\\ =\int e^t\left[f\left(x\right)+f^{\prime }\left(x\right)\right]x\\ =e^{t}f\left(x\right)+C\\ =e^t\frac{1}{1+t}+C\\ =\frac{x}{1+\log x}+C\end{array}$