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

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Answer 1 · 2018-06-03T05:39:54

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

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

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

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

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