Answer 1

Problem:

`intlogx/(1+logx)^2dx`
adding and substracting 1 from numerator

`int (1-1+logx)/(1+logx)^2dx`

`int (1+logx)/(1+logx)^2dx-int(1)/(1+logx)^2 dx`

`int 1/(1+logx)dx-int(1)/(1+logx)^2 dx`
For the integral

` int 1/(1+logx)dx`
integrate by parts within the sum: ∫fg'=fg−∫f'g

`f= 1/(1+logx)dx, g'=1`

`f'=-(1)/(1+logx)^2, g=x`

`=-int(1)/(1+logx)^2 dx-int -1/(1+logx)^2dx+x/(log(x)+1)`

`=x/(log(x)+1)`