How do you find the Integral of $dx/sqrt(x^2+16)$ ?

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Answer 1 · 2015-03-30T17:04:07

$int (dx)/ sqrt(x^2+16) dx = int 1/(4sqrt((x/4)^2+1)) dx$

$=int 1/(sqrt((x/4)^2+1))( 1/4)dx$

$u = x/4$ , $du=1/4 dx$ and $int 1/sqrt(u^2+1) du=sinh^(-1) u +C$

So
$int (dx)/ sqrt(x^2+16) dx = sinh^(-1) (x/4) +C$

How do you find the Integral of dx/sqrt(x^2+16)dx/sqrt(x^2+16)?