Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$. 
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$


Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$  Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \,  du\\
 &= \frac1{16}\int {\sec^4 u} \, du\\ 
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
 &= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C. 
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
 A: Hint:
$$\tan(\sec^{-1}(x))=\pm\sqrt{x^2-1}$$
(that is, your answer simplifies a bit further)
A: Once you get to $\tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C$, get rid of $u$. Since $\sec u = 2x = 2x/1$, that means $\tan u = \sqrt{4x^2 - 1}$ (draw a right triangle to show this), so $$\begin{aligned}[t]\text{integral} = \tfrac{1}{16}(\tan u + \tfrac{1}{3} \tan^3 u) + C &= \tfrac{1}{16}\Big(\sqrt{4x^2 - 1} + \tfrac{1}{3} (4x^2-1)\sqrt{4x^2-1}\,\Bigr) + C \\ &= \tfrac{1}{16}\sqrt{4x^2-1} \, \Bigl(1 + \tfrac{4}{3} x^2 - \tfrac{1}{3} \Bigr) + C \\ &= \tfrac{1}{24}\sqrt{4x^2-1} \, \Bigl(2x^2+1 \Bigr) + C.\end{aligned}$$
A: $$\tan\left(\sec^{-1}(2x)\right)=\sqrt{4x^2-1}$$
Make this substitution into your answer and you will get the result you expected.
A: There is nothing wrong with your approach; you just need to fiddle around with trig identities to see that $\tan(\sec^{-1}2x)=\sqrt{\sec^2(\sec^{-1}2x)-1}=\sqrt{(2x)^2-1}$, etc.  But another, possibly easier, way to do the integral is to let $u=4x^2$ so that $du=8x\,dx$ and thus
$$\int{x^3\over\sqrt{4x^2-1}}dx={1\over32}\int{u\over\sqrt{u-1}}du={1\over32}\int\left(\sqrt{u-1}+{1\over\sqrt{u-1}}\right)du\\
={1\over48}(u-1)^{3/2}+{1\over16}(u-1)^{1/2}+C=\sqrt{u-1}\left(u+2\over48\right)+C\\
=\sqrt{4x^2-1}\left(4x^2+2\over48\right)+C=\sqrt{4x^2-1}\left(2x^2+1\over24\right)+C$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{t \equiv \root{4x^{2} - 1} - 2x\ \mbox{and}\ x = -\,{t^{2} + 1 \over 4t}}$:
\begin{align}
\int{x^3 \over \root{4x^{2} - 1}}\,\dd x & =
{1 \over 128}\int{\pars{t^{2} + 1}^{3} \over t^{4}}\,\dd t
\end{align}

Another one is
  $\ds{y = x^{2} \implies {1 \over 2}\int{y \over \root{4y - 1}}\,\dd y}$.

A: 
By the above diagram, we can simplify the answer obtained by the author.
$$
\begin{array}{l}
\begin{aligned}
I &=\frac{1}{16}\left(\tan \theta+\frac{\tan ^{3} \theta}{3}\right)+C \\
&=\frac{1}{16}\left[\sqrt{4 x^{2}-1}+\frac{\left(4 x^{2}-1\right)^{\frac{3}{2}}}{3}\right]+C
\\&=\frac{\sqrt{4 x^{2}-1}}{48}\left[3+4 x^{2}-1\right]+C \\
\\&=\frac{\left(2 x^{2}+1\right) \sqrt{4 x^{2}-1}}{24}+C
\end{aligned} \\
\end{array}
$$
