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.
 A: Continue from what you have already figured out: $$\int\frac{x^{2}}{\sqrt{x^{2}-16}}dx=16\int\tan^{2}(u)\sec (u)\ du$$
Recall $\tan^2(x) = \sec^2(x)-1$,$$\int\tan^2(u)\sec(u) du=\int\sec^3(u) du - \int\sec(u) du$$
Then using $\int\sec^3(u)\ du = \frac12\tan(u)\sec(u) + \frac12\ln|\tan(u) + \sec(u)| + C$ (integration by parts) and $\int\sec(u)\ du = \ln|\tan(x) + \sec(u)| + C$ to solve the problem.
A: Take $x=4\sec\left(u\right)
 $. We get $$I=\int\frac{x^{2}}{\sqrt{x^{2}-16}}dx=16\int\sec^{3}\left(u\right)du
 $$ and now for the reduction formula for powers of $\sec\left(u\right)
 $ we have $$I=8\tan\left(u\right)\sec\left(u\right)+8\int\sec\left(u\right)du
 $$ $$=8\tan\left(u\right)\sec\left(u\right)+8\log\left(\tan\left(u\right)+\sec\left(u\right)\right)
 $$ and now substitute back.
A: First of all do $x=4t$, so you reduce to
$$
64\int\frac{t^2}{\sqrt{t^2-1}}\,dt
$$
Leave the coefficient alone to reinsert it later. Now substitute $t=\cosh u$, with $u>0$, so $\sqrt{t^2-1}=\sqrt{\cosh^2u-1}=\sinh u$ and $dt=\sinh u\,du$, so the integral is
$$
\int\cosh^2u\,du=\int\cosh u\cdot\cosh u\,du
$$
Use integration by parts:
\begin{align}
\int\cosh u\cdot\cosh u\,du
&=\cosh u\sinh u-\int\sinh^2u\,du\\
&=\cosh u\sinh u-\int(\cosh^2u-1)\,du\\
&=\cosh u\sinh u+u-\int\cosh^2\,du
\end{align}
so
$$
\int\cosh^2u\,du=\frac{1}{2}u+\frac{1}{2}\cosh u\sinh u+C
$$
Back substitute, reinsert $64$ and you're done.
A: 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
$$\int \frac{x^2}{\sqrt{x^2-16}} dx = \frac{1}{2} x \sqrt{16-x^2} + 8 \ln\left(\sqrt{x^2-16}+x\right)+C$$
A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{\quad t \equiv x - \root{x^{2} - 16}\quad\imp\quad
x = {t^{2} + 16 \over 2t}}$:
\begin{align}
\color{#f00}{\int{x^{2} \over \root{x^{2} - 16}}\,\dd x} & =
\int\pars{-\,{64 \over t^{3}} - {8 \over t} - {t \over 4}}\,\dd t =
{32 \over t^{2}} - 8\ln\pars{t} - {1 \over 8}\,t^{2}
\\[3mm] & =
{32 \over \pars{x - \root{x^{2} - 16}}^{2}} -
8\ln\pars{x - \root{x^{2} - 16}} - {1 \over 8}\pars{x - \root{x^{2} - 16}}^{2}
\\[3mm] & =
{1 \over 8}\,\bracks{\pars{x + \root{x^{2} - 16}}^{2} -
\pars{x - \root{x^{2} - 16}}^{2}} - 8\ln\pars{x - \root{x^{2} - 16}}
\\[3mm] & =
\color{#f00}{\half\,x\root{x^{2} - 16} - 8\ln\pars{x - \root{x^{2} - 16}}}
\end{align}
