`int(1+logx)/(x(2x+log x)(3+logx))dx` (March '16)

Explanation

Let `I =int(1+logx)/(x(2+logx)(3+logx))dx`
Put `1+logx+timplies1/xdx=dt`
`I=int (t)/((1+t)(2+t))dt" "...(1)`
Let `(t)/((1+t)(2+t))=(A)/(1+t)+(B)/(2+t)`
`t=A(2+t)+B(1+t)" "...(2)`
Putting `t=-1` in equation (2), we get
`-1 =A`
`impliesA=-1`
Putting `t=-2` in equation (2), we get
`-2=-B`
`impliesB=2`
Then equation (1), becomes
`I=int[(-1)/(1+t)+(2)/(2+t)]dt`
`=-int(1)/(1+t)dt+2int(1)/(2+t)dt`
`=-log|t+1|+2log|t+2log|t+2|+c`
`=2log|logx+3|-log|logx+2|+c`
`=log|(logx+3)^(2)|-log|(logx+2)|+c`
`therefore int(1+logx)/(x(2+logx)(3+logx))dx`
`=log|((logx+3)^(2))/((logx+2))|+x`