How do you integrate $int x^3/sqrt(x^2+9)$ using trig substitutions?

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Answer 1 · 2016-12-17T10:46:59

$=1/3sqrt(x^2+9)( x^2-18)+C$

Let $t = x^2+9$ .

Then, $dt =2x dx$ .

Now, the given integral becomes

$1/2int (t-9)/sqrt t dt$

$=1/2int(t^(1/2)-9t^(-1/2)) dt$

$=1/2(2/3t^(3/2)-18t^(1/2))+C$

$=1/3sqrt t(t-27)+C$

$=1/3sqrt(x^2+9)( x^2-18)+C$ -

How do you integrate int x^3/sqrt(x^2+9)int x^3/sqrt(x^2+9) using trig substitutions?