<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] Note that
\begin{align}
\int \frac{x^2}{\sqrt{x^2-16}} dx &= \int \frac{(x^2-16)+16}{\sqrt{x^2-16}}dx \\
&=\int \sqrt{x^2-16}dx + 16 \int \frac{dx}{\sqrt{x^2-16}} \\
&= \sqrt{x^2-16} x - \int \frac{x^2}{\sqrt{x^2-16}}dx+ 16 \int \frac{dx}{\sqrt{x^2-16}}dx
\end{align}
Hence
$$\int \frac{x^2}{\sqrt{x^2-16}} = \frac{1}{2} x \sqrt{16-x^2} + 8 \int \frac{dx}{\sqrt{x^2-16}}dx. $$
One verifies that
$$\int \frac{dx}{\sqrt{x^2-16}} = \ln\left(\sqrt{x^2-16}+x\right)$$
Therefore
$$\int \frac{x^2}{\sqrt{x^2-16}} dx = \frac{1}{2} x \sqrt{16-x^2} + 8 \ln\left(\sqrt{x^2-16}+x\right)+C$$
</s>