Integrate the functions`(sqrt(x^2+1)[log(x^2+1)-2logx])/(x^4)`

Explanation

Let
`I= int (sqrt(x^(2)+1)[log (x^(2)+1)-2log x])/(x^(4))dx`
`=int sqrt((x^(2)+1)/(x^(4)))[log (x^(2)+1)-log x^(2)]dx`
`=int (sqrt(x^(2)+1))/(x^(4))[log (1+(1)/(x^(2)))]dx`
`=int (1)/(x^(3))sqrt((x^(2)+1)/(x^(2)))[log (1+(1)/(x^(2)))]dx`
`underset(rArr" "(dx)/(x^(3)) =-(1)/(2)dt)underset(rArr" "1+(1)/(x^(2))=trArr -(2)/(x^(3))dx)("Let " (x^(2)+1)/(x^(2))=t)`
`:.I= int sqrt(t) log t. (-(dt)/(2))=-(1)/(2) int (t)^(1//2) log t dt`
`=-(1)/(2)[log t.int t^(1//2)dt-int {((d)/(dx) log t )int t^(1//2)dt}dt]`
`=-(1)/(2) [log t.(t^(3//2))/(3//2)-int (1)/(t).(t^(3//2))/((3)/(2))dt]`
`=-(1)/(2)[(2)/(3) log t. t^(3//2) -(2)/(3)int t^(1//2)dt]`
`=-(1)/(2)xx(2)/(3) [log t.t^(3//2)-int t^(1//2) dt]`
`=-(1)/(3)(log t.t^(3//2)-(t^(3//2))/(3//2))+C`
`=-(1)/(3)t^(3//2) [log t-(2)/(3)]+C`
`=-(1)/(3) (1+(1)/(x^(2)))^(3//2)[log ((x^(2)+1)/(x^(2)))-(2)/(3)]+C`