Question

How do you evaluate the integral `int dx/(x^2sqrt(x^2-3))`?

Answer 1

After using `x=sqrt(3)*secu` and `dx=sqrt(3)*secu*tanu*du` transforms, this integral became,

`int (sqrt(3)*secu*tanu*(du))/[3(secu)^2*sqrt(3)*tanu]`

=`1/3*int (du)/secu`

=`1/3*int cosu*du`

=`1/3*sinu+C`

After using `x=sqrt(3)*secu`, `secu=x/sqrt(3)`, `tanu=sqrt(x^2-3)/sqrt(3)` and `sinu=tanu/secu=sqrt(x^2-3)/x` inverse transforms, I found,

`int (dx)/[x^2*sqrt(x^2-3)]=sqrt(x^2-3)/(3x)+C`

Explanation:

I used `x=sqrt(3)*secu` and `dx=sqrt(3)*secu*tanu*du` transform