Question

How do you integrate `int 1/sqrt(4x^2+16x-5)` using trigonometric substitution?

Answer 1

`-1/8*ln|t-4|+C`
where `t=sqrt(4x^2+16x-5)-2x`

Explanation:

Setting
`sqrt(4x^2+16x-5)=2x+t`
then we get (by squaring)

`x=(5+t^2)/(16-4t)`
`dx=(-4t^2+32t+20)/(16-4t)^2dt`
so

`sqrt(4x^2+16x-5)=2(5+t^2)/(16-4t)+t=(-t^2+8t+5)/(8-2t)`
and our integral will be

`int (8-2t)/(-t^2+8t+5)*4(-t^2+8t+5)/(16-4t)^2dt`

`-1/8int dt/(t-4)=-1/8*ln|t-4|+C`