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

Question

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Answer 1 · 2016-12-17T10:46:59

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

Let $t = x^{2} + 9$ $t = x^2+9$ .

Then, $d t = 2 x d x$ $dt =2x dx$ .

Now, the given integral becomes

$\frac{1}{2} \int \frac{t - 9}{\sqrt{t}} d t$ $1/2int (t-9)/sqrt t dt$

$= \frac{1}{2} \int (t^{\frac{1}{2}} - 9 t^{- \frac{1}{2}}) d t$ $=1/2int(t^(1/2)-9t^(-1/2)) dt$

$= \frac{1}{2} (\frac{2}{3} t^{\frac{3}{2}} - 18 t^{\frac{1}{2}}) + C$ $=1/2(2/3t^(3/2)-18t^(1/2))+C$

$= \frac{1}{3} \sqrt{t} (t - 27) + C$ $=1/3sqrt t(t-27)+C$

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

How do you integrate ∫x3√x2+9int x^3/sqrt(x^2+9) using trig substitutions?