<s>[INST] 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?
 [/INST] 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}
</s>