Integrating $\int \frac{u \,du}{(a^2+u^2)^{3/2}}$ How does one integrate $$\int \frac{u \,du}{(a^2+u^2)^{3/2}} ?$$
Looking at it, the substitution rule seems like method of choice. What is the strategy here for choosing a substitution?
 A: $$\int\frac{u}{\left(a^2+u^2\right)^{\frac{3}{2}}}\space\text{d}u=$$

Substitute $s=a^2+u^2$ and $\text{d}s=2u\space\text{d}u$:

$$\frac{1}{2}\int\frac{1}{s^{\frac{3}{2}}}\space\text{d}s=\frac{1}{2}\int s^{-\frac{3}{2}}\space\text{d}s=\frac{1}{2}\cdot-\frac{2}{\sqrt{s}}+\text{C}=-\frac{1}{\sqrt{a^2+u^2}}+\text{C}$$
A: Let $w=$ some function of $u$ for which $dw = u\,du$.
A: Besides the other answers, I would like to expose another way of doing it:
$$\int \dfrac{x}{(a^2+x^2)^{3/2}} dx=-2\int \dfrac{1}{(a^2+x^2)^{1/4}}\cdot \dfrac {\dfrac{-x}{2} (a^2+x^2)^{-3/4}}{(a^2+x^2)^{1/2}} \, dx$$ We realise now that the second factor is the derivative of the first, so we use $f(x)=x$ and $g(x)=\dfrac{1}{(a^2+x^2)^{1/4}}$, to get $\int u$ $du= \dfrac{u^2}{2}$. Don't forget to multiply by the -2, by which we get  $-u^2=-\dfrac{1}{(a^2+x^2)^{1/2}}$.  Although I think this way to proceed is more intrincate than the other answers, it has the advantage of being generalizable:
If $R, l \in \mathbb{R}$, $$\int \dfrac{x}{(R+x^2)^l} \, dx=\int \dfrac{1}{(R+x^2)^p}\cdot \dfrac{x}{(R+x^2)^{p+1}} \, dx$$ (here $p=\dfrac{l-1}{2}$)
$$=\dfrac{-1}{2p} \int \dfrac{1}{(R+x^2)^{p}} \cdot \dfrac{-2px(R+x^2)^{p-1}}{(R+x^2)^{2p}} \, dx =\dfrac{-1}{2p} \cdot \dfrac{u^2}{2}=\dfrac{-1}{4p (R+x^2)^{2p}}$$ Recalling that $l=2p+1$, we finally get: $\dfrac{1}{(2-2l)(R+x^2)^{l-1}} + \text{Constant}$.
Edit: I changed the $u$ of the OP by $x$ only for a matter of comfort.
A: You can take the solution as I've presented here:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) -  B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n}  \right)\right]
\end{equation}
Here $a$ is $a^2$, $k = 1$, $m = \frac{3}{2}$, thus:
\begin{align}
\int_0^x \frac{t}{\left(t^{2} + a^{2}\right)^{\frac{3}{2}}}\:dt &= \frac{1}{2}\left(a^2\right)^{\frac{1 + 1}{2} - \frac{3}{2}} \left[B\left(\frac{3}{2} - \frac{1 + 1}{2}, \frac{1 + 1}{2}\right) -  B\left(\frac{3}{2} - \frac{1 + 1}{2}, \frac{1 + 1}{2}, \frac{1}{1 + a^2x^2}  \right)\right] \\
&= \frac{1}{\left|a\right|}\left[B\left(\frac{1}{2}, 1\right) - B\left(\frac{1}{2},1,  \frac{1}{1 + a^2x^2} \right) \right]
\end{align}
Using the relationship between the Beta and the Gamma function we find that:
\begin{equation}
 B\left(\frac{1}{2},1 \right) =  \frac{\Gamma\left(\frac{1}{2} \right)\Gamma(1)}{\Gamma\left(1 + \frac{3}{2}\right)}  = \frac{\Gamma\left(\frac{1}{2} \right)\Gamma(1)}{\frac{1}{2}\Gamma\left(\frac{1}{2}\right)} = 2\Gamma(1) = 2
\end{equation}
Thus, 
\begin{equation}
 \int_0^x \frac{t}{\left(t^{2} + a^{2}\right)^{\frac{3}{2}}}\:dt = \frac{1}{\left|a\right|}\left[2 - B\left(\frac{1}{2},1,  \frac{1}{1 + a^2x^2} \right) \right]
\end{equation}
