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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?

Travis Willse
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Weverton Alison
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    The first one that comes to my mind is $t=a^2+u^2$. Did you try it? – mickep Jan 15 '16 at 18:48
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    A clumsier one that comes to mind is $t=u^2$, because the $u,du$ is going to pop out when you relate $du$ to $dt$, and because $u^2$ appears in the complicated part of the integrand and it might be easier if that was the variable of integration. This substitution works as well, although @mickep's suggestion is more elegant. – Mark Fischler Jan 15 '16 at 18:51
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    if you set $t=a^2+u^2$ you will get $dt=2udu$ – Dr. Sonnhard Graubner Jan 15 '16 at 18:53
  • $\partial_u \frac{1}{(u^2+a^2)^{1/2}}= -\frac{u}{(u^2+a^2)^{3/2}}$ – tired Jan 15 '16 at 19:38

4 Answers4

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

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

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

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

Michael Hardy
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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.

Michael Hardy
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Diego
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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}