Question

What is `int 23/sqrt(13+3x^2) dx`?

Answer 1

`23int1/sqrt(13+3x^2)dx`

`x = sqrt(13/3)u`

`dx = sqrt(13/3)du`

`u = sqrt(3/13)x`

`23sqrt(13/3)int 1/sqrt(13+13u^2)du`

`23sqrt(13/3)*sqrt(1/13)int 1/sqrt(1+u^2)du`

`23sqrt(1/3)[arcsinh(u)]+C`

Substitute back

`23sqrt(1/3)[arcsinh(sqrt(3/13)x)]+C`