Integrating $\int \frac{u \,du}{(a^2+u^2)^{3/2}}$

Question

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?

Answer 1

$$\int\frac{u}{\left(a^2+u^2\right)^{\frac{3}{2}}}\space\text{d}u=$$

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

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$$\frac{1}{2}\int\frac{1}{s^{\frac{3}{2}}}\space\text{d}s=\frac{1}{2}\int s^{-\frac{3}{2}}\space\text{d}s=\frac{1}{2}\cdot-\frac{2}{\sqrt{s}}+\text{C}=-\frac{1}{\sqrt{a^2+u^2}}+\text{C}$$

Answer 2

Let $w=$ some function of $u$ for which $dw = u\,du$.

Answer 3

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.