Question

How do you find the integral `int1/(x^2*sqrt(x^2-9))dx` ?

Answer 1

`=1/(9x)*sqrt(x^2-9)+c`, where `c` is a constant

Explanation :

`=int1/(x^2sqrt(x^2-9))dx`

`=int1/(x^3sqrt(1-9/x^2))dx`

Using Integration by Substitution,

let's assume `3/x=t`, `-3/x^2dx=dt`

`=int-t/(9sqrt(1-t^2))dt=-1/9intt/(sqrt(1-t^2))dt`

again assuming `t^2=z`, `2tdt=dz`

`=-1/9int1/2*dz/sqrt(1-z)`

`=-1/18int1/sqrt(1-z)dz`

`=-1/18int(1-z)^(-1/2)dz`

`=1/18*(1-z)^(1/2)/(1/2)+c`, where `c` is a constant

Substituting value of `z` and `t` back,

`=1/9sqrt(1-9/x^2)+c`, where `c` is a constant

`=1/(9x)*sqrt(x^2-9)+c`, where `c` is a constant