Question

Evaluate: 

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

Answer 1

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

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

Put log x = t

∴ `1/x  dx = dt`

∴`int(1+t)/((3+t)(2+3t))  dt`

Let `(1+t)/((3+t)(2+3t)) = A/(3+t) + B/(2+3t)`

∴ `1 + t = A(2+3t)+ B(3+t)`      ...(1)

Put `3+ t = 0`, i.e. `t = -3` in (1), we get

`1-3 = A(-7) + B(0)`

∴ `-2=-7A`

∴ `A= 2/7`

Put `2+ 3t = 0`, i.e. `t = -2/3` in (1), we get

`1-2/3 = A(0)+ B(7/3)`

∴ `1/3 = 7/3 B`

∴ `B=1/7`

∴ `(1+t)/((3+t)(2+3t)) = (2/7)/(3+t)+(1/7)/(2+3t)`

∴ I = `int[(2/7)/(3+t)+(1/7)/(2+3t)]dt`

=`2/7int1/(3+t) dt + 1/7 int1/(2+3t)dt`

`=2/7log|3+t|+1/7. (log|2+3t|)/3 + c`

`2/7 log |3+logx|+1/21 log|2+3logx|+c`