Question

How do you find the integral of `int 1/(xsqrt(x^4-4)`?

Answer 1

`intdx/(xsqrt(x^4-4))=1/4sec^-1(x^2/2)+C`

Explanation:

`I=intdx/(xsqrt(x^4-4))`

Try the substitution `x^2=2sectheta`. This implies that `2xdx=2secthetatanthetad theta`.

It also implies that `sqrt(x^4-4)=sqrt(4sec^2theta-4)=2sqrt(sec^2theta-1)=2tantheta`. Then:

`I=1/2int(2xdx)/(x^2sqrt(x^4-4))`

`I=1/2int(2secthetatanthetad theta)/(2sectheta(2tantheta))`

`I=1/4intd theta`

`I=1/4theta`

From `x^2=2sectheta` we see that `theta=sec^-1(x^2/2)`:

`I=1/4sec^-1(x^2/2)+C`