Question

Evaluate: $\int \frac{\log x}{(1+\log x{)}^2}dx$

Answer 1

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

adding and subtracting 1 from numerator

$\int \frac{1-1+logx}{(1+logx{)}^2}dx$

$\int \frac{1+logx}{(1+logx{)}^2}dx-\int \frac{1}{(1+logx{)}^2}dx$

$\int \frac{1}{1+logx}dx-\int {\frac{1}{(1+logx)}}^{2}dx$

For the integral

$\int \frac{1}{1+logx}dx$

integrate by parts within the sum : $\int f{g}^{\prime }=fg-\int f^{\prime }g$

$f=\frac{1}{1+logx}dx,g^{\prime }=1$

$f^{\prime }=-\frac{1}{(1+logx{)}^2}g=x$

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

$=\frac{x}{log\left(x\right)+1}$