Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?
 A: Here's one way to go about deriving a recursion relation for integrals of the form
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx$$
Split the integral like so:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx$$
and integrate by parts:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\sqrt{1+x^2} x^{n-2}\mathrm dx$$
Noting that $1+x^2$ is always positive for real $x$, we then complicate things a little:
$$\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int(1+x^2)\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
Perform another split:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\left(\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx+\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
and we see something we can isolate:
$$n\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
and then we finally divide both sides by $n$:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=\frac1{n}\left(x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
We can use the starting values $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\mathrm{arsinh}\,x$ and $\int\frac{x \mathrm dx}{\sqrt{1+x^2}}=\sqrt{1+x^2}$ for the recursion.
(This is a response to Srivatsan's comment, which got too long for the comment box.)
A: I would first try the substitution $x=\tan(\theta)$, so that $\sqrt{1+x^2}=\sec(\theta)$. That gives
$$
\begin{align}
\int\frac{x^n}{\sqrt{1+x^2}}\;\mathrm{d}x
&=\int \tan^n(\theta)\sec(\theta)\;\mathrm{d}\theta\\
&=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int\tan^{n-2}(\theta)\;\sec^3(\theta)\;\mathrm{d}\theta\\
&=\tan^{n-1}(\theta)\sec(\theta)-(n-1)\int(\tan^n(\theta)+\tan^{n-2}(\theta))\;\sec(\theta)\;\mathrm{d}\theta\\
&=\frac{1}{n}\tan^{n-1}(\theta)\sec(\theta)-\frac{n-1}{n}\int\tan^{n-2}(\theta)\;\sec(\theta)\;\mathrm{d}\theta
\end{align}
$$
If $n$ is odd, this reduces to
$$
\int\tan(\theta)\sec(\theta)\;\mathrm{d}\theta=\sec(\theta)+C
$$
If $n$ is even, this reduces to 
$$
\begin{align}
\int\sec(\theta)\;\mathrm{d}\theta&=\int\sec^2(\theta)\;\mathrm{d}\sin(\theta)\\
&=\int\frac{1}{2}\left(\frac{1}{1-\sin(\theta)}+\frac{1}{1+\sin(\theta)}\right)\;\mathrm{d}\sin(\theta)\\
&=\frac{1}{2}\log\left(\frac{1+\sin(\theta)}{1-\sin(\theta)}\right)+C\\
&=\log(\sec(\theta)+\tan(\theta))+C
\end{align}
$$
A: Recall the hyperbolic functions 
$$\cosh t= \frac{e^t + e^{-t}}{2} = \cos(it)$$ 
and $$\sinh t=\frac{e^t - e^{-t}}{2} = i\sin(-it).$$
Note that $\frac{d}{dt}\sinh t = \cosh t$, $\frac{d}{dt}\cosh t = \sinh t$ and also $\cosh^2 t -\sinh^2 t = 1$.
Making the substitution $\sinh t=x $ we see that
$$\frac{x^n\, dx}{\sqrt{1+x^2}} = \frac{\sinh^n t\, \cosh t\,dt}{\sqrt{1+\sinh^2t}}=
\frac{\sinh^n t\, \cosh t\,dt}{\sqrt{\cosh^2t}}=\sinh^n t\, dt$$
which leads us to $$\int\frac{x^n\, dx}{\sqrt{1+x^2}} = \int \sinh^n t\, dt.$$
To complete the problem, the binomial theorem is useful.
A: Since $\frac{d}{dt}\sqrt{1+t^2} = \frac{t}{\sqrt{1+t^2}}$, we can integrate by parts to get
$$
\int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt = \int t\cdot \frac{t}{\sqrt{1+t^2}}\mathrm dt
= t\sqrt{1+t^2} - \int \sqrt{1+t^2}\mathrm dt.
$$
Cheating a little bit by looking at a table of integrals, we get that since
$$
\frac{d}{dt} \left [ t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2}) \right ]
= t\frac{t}{\sqrt{1+t^2}} + \sqrt{1+t^2} + \frac{1}{t + \sqrt{1+t^2}}
\left [ 1 + \frac{t}{\sqrt{1+t^2}} \right ]
$$
which simplifies to $2\sqrt{1+t^2}$, the integral on the right above is 
$\frac{1}{2}[t\sqrt{1+t^2} + \ln(t + \sqrt{1+t^2})]$ and thus we have
$$
\int \frac{t^2}{\sqrt{1+t^2}}\mathrm dt 
= \frac{1}{2}\left [t\sqrt{1+t^2} - \ln(t + \sqrt{1+t^2})\right ] 
$$
which matches the answer given by Charles Bao if we set $X=x$ and $a=1$ in his original post.
