Use substitution techniques and a table of integrals to find the indefinite integral.
\[
\int \frac{x}{\sqrt{x^{4}-16}} d x
\]
$\exists$ Click the icon to view a brief table of integrals.
Choose the most useful substitution below.
A. $u=\sqrt{x^{4}-16}$
B. $u=x$
C. $u=x^{4}$
D. $u=x^{4}-16$
E. $u=x^{2}$
\[
\int \frac{x}{\sqrt{x^{4}-16}} d x=2 x+C
\]

Step 1 ：Given the integral \(\int \frac{x}{\sqrt{x^{4}-16}} dx\)

Step 2 ：Choose the substitution $u = x^4 - 16$

Step 3 ：After the substitution, the integral becomes a standard form that can be solved directly

Step 4 ：The integral after substitution and simplification is \(\frac{1}{2} \cosh ^{-1}\left(\frac{x^{2}}{4}\right)\) when \(|x^{4}|>16\) and \(-\frac{i}{2} \sin ^{-1}\left(\frac{x^{2}}{4}\right)\) otherwise

Step 5 ：Thus, the indefinite integral of \(\frac{x}{\sqrt{x^{4}-16}}\) with respect to $x$ is \[\begin{cases} \frac{1}{2} \cosh ^{-1}\left(\frac{x^{2}}{4}\right)+C & \text{if } |x^{4}|>16 \\ -\frac{i}{2} \sin ^{-1}\left(\frac{x^{2}}{4}\right)+C & \text{otherwise} \end{cases}\] where $C$ is the constant of integration

Step 6 ：\(\boxed{\begin{cases} \frac{1}{2} \cosh ^{-1}\left(\frac{x^{2}}{4}\right)+C & \text{if } |x^{4}|>16 \\ -\frac{i}{2} \sin ^{-1}\left(\frac{x^{2}}{4}\right)+C & \text{otherwise} \end{cases}}\)