How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
 A: You made a small mistake on the last line; integrating $\frac{1}{2}u^{1/2}-\frac{1}{2}u^{-1/2}$ should give $\frac{1}{3}u^{3/2}-u^{1/2}+C$, not $\frac{2}{3}u^{3/2}-\frac{1}{2}u^{1/2}+C$. When you want to double-check an antiderivative, it's worth differentiating it to see if you get the expected integrand.
A: Here I give you an alternative approach. If we make $x = \sinh(u)$, we get
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x & = \int\sinh^{3}(u)\mathrm{d}u = \int(\cosh^{2}(u)-1)\sinh(u)\mathrm{d}u\\
& = \frac{\cosh^{3}(u)}{3} - \cosh(u) + K
\end{align*}
Since $x = \sinh(u) = \pm\sqrt{\cosh^{2}(u)-1}$, we conclude that $\cosh(u) = \sqrt{x^{2} + 1}$. Finally, it results that
\begin{align*}
\int\frac{x^{3}}{\sqrt{x^{2}+1}}\mathrm{d}x = \frac{(x^{2}+1)^{3/2}}{3} - \sqrt{x^{2}+1} + K
\end{align*}
A: Just to give another approach, let $x^2+1=u^2$, in which case $xdx=udu$ and we have
$$\int{x^3\over\sqrt{x^2+1}}dx=\int{x^2\cdot xdx\over\sqrt{x^2+1}}=\int{(u^2-1)udu\over u}=\int(u^2-1)du={1\over3}u^3-u+C\\={1\over3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C$$
The main virtue (if any) of this approach is that it gets rid of the square root symbol for the integration step.
A: Method 1:
Another approach using trigonometric substitutions
\begin{equation}
I =\int \frac{x^3}{\sqrt {x^2+1}}dx
\end{equation}
Here let $x = \tan(\theta)$ we arrive at:
\begin{align}
I &=\int \frac{\tan^3(\theta)}{\sqrt {\tan^2(\theta)+1}}\sec^2(\theta)\:d\theta = \int \frac{\tan^3(\theta)}{\sec(\theta)}\sec^2(\theta)\:d\theta\\
&= \int \tan^3(\theta)\sec(\theta)\:d\theta = \int \sec(\theta)\tan(\theta) tan^2(\theta)\:d\theta \\
&= \int \sec(\theta)\tan(\theta) \left(\sec^2(\theta) - 1\right)\:d\theta
\end{align}
Now let $u = \sec(\theta)$ to yield:
\begin{equation}
  \int \sec(\theta)\tan(\theta)\left(u^2 - 1\right) \frac{\:d\theta}{\sec(\theta)\tan(\theta)} =  \int u^2 - 1 \:du = \frac{u^3}{3} - u + C
\end{equation}
Where $C$ is the constant of integration. 
Now $u = \sec(\theta)$ and $x = \tan(\theta)$ and so $u = \sec(\arctan(x)) = \sqrt{x^2 + 1}$
Thus, 
\begin{equation}
I =\int \frac{x^3}{\sqrt {x^2+1}}dx = \frac{1}{3}\left( \sqrt {x^2+1}\right)^3  - \sqrt {x^2+1}  + C =  \frac{1}{3}\left(x^2+1\right)^{\frac{3}{2}}  - \sqrt {x^2+1}  + C
\end{equation}
Method 2:
\begin{align}
I &=\int \frac{x^3}{\sqrt {x^2+1}}dx = \int x \cdot \frac{x^2}{\sqrt {x^2+1}}dx \\
 &= \int x \cdot \frac{x^2 + 1 - 1}{\sqrt {x^2+1}}dx = \int x \left( \sqrt {x^2+1} - \frac{1}{\sqrt{x^2+1}}\right)dx
\end{align}
Here let $u = x^2 + 1$:
\begin{align}
 I &= \int x \left( \sqrt {u} - \frac{1}{\sqrt{u}}\right)\frac{du}{2x} = \frac{1}{2}\left[\int \left( u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du\right] = \frac{1}{2}\left[\frac{ u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{ u^{\frac{1}{2}}}{\frac{1}{2}}  \right] + C \\
&= \frac{1}{3}u^{\frac{3}{2}} + u^{\frac{1}{2}} + C = \frac{1}{3}\left(x^2 + 1\right)^{\frac{3}{2}} + \left(x^2 + 1\right)^{\frac{1}{2}} + C 
\end{align}
Where $C$ is the constant of integration.
