<s>[INST] Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort 2:
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}$
Let's doing integral by parts;
$du=\dfrac{x}{\sqrt{x^2-16}}$
$v=x$
$\displaystyle\int \dfrac {x^{2}dx} {\sqrt {x^{2}-16}}=x.\sqrt{x^2-16}-\displaystyle\int\sqrt{x^2-16}dx$
We have $\quad \displaystyle\int\sqrt{x^2-16}dx$
let be $\quad x=4\sec j$
$dx=4\dfrac{\sin j}{\cos^2 j}dj$
and;
$\displaystyle\int\sqrt{x^2-16}\;dx=16.\displaystyle\int\dfrac{\sin j}{\cos j}\dfrac{\sin j}{\cos^2 j}dj=16.\displaystyle\int \sec j.tan^2j.dj$
After I didn't nothing.
 [/INST] Just another way to compute the antiderivative.
Change variable $$x^2-16=t^2\implies x=\sqrt{t^2+16}\implies dx=\frac{t}{\sqrt{t^2+16}}\,dt$$
So, $$I=\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx=\int \sqrt{t^2+16}\,dt$$ One integration by parts $$u=\sqrt{t^2+16}\implies du=\frac{t}{\sqrt{t^2+16}}\,dt$$ $dv=dt \implies v=t$. So $$I=\int \sqrt{t^2+16}\,dt=t \sqrt{t^2+16}-\int \frac{t^2}{\sqrt{t^2+16}}\,dt$$ $$I=t \sqrt{t^2+16}-\int \frac{t^2+16-16}{\sqrt{t^2+16}}\,dt=\sqrt{t^2+16}-I+\int \frac{16}{\sqrt{t^2+16}}\,dt$$ which makes $$2I=t \sqrt{t^2+16}+\int \frac{16}{\sqrt{t^2+16}}\,dt$$ Now, an obvious change of variable $t=4\sinh(u)$ will give the end result easily.
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