Answer 1

`int(1+logx)/(x(2+logx)(3+logx))dx`

Substitute logx = t..................(1)

`therefore 1/xdx=dt`

Hence, the integral becomes

`int(1+t)/((2+t)(3+t))dt`

`=int(2+t-1)/((2+t)(3+t))dt`

`=int(2+t)/((2+t)(3+t))dt-int1/((2+t)(3+t))dt`

`=int1/(3+t)dt-int((t+3)-(t+2))/((2+t)(3+t))dt`

`=int1/(3+t)dt-[int(t+3)/((2+t)(3+t))dt-int(t+2)/((2+t)(3+t))dt]`

`=int1/(3+t)dt-int1/(2+t)dt+int1/(3+t)dt`

`=2int1/(3+t)dt-int1/(2+t)dt`

`=2int1/(3+t)dt-int1/(2+t)dt`

Substituting the value of 't' from (1), we get

`int(1+logx)/(x(2+logx)(3+logx))dx`

`2ln (3+ logx )-ln( 2+ logx)+ C`

`=log|(3+logx)^2/(2 + logx)| + C`