Cfrgtkky

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?

$$intfrac{u}{left(a^2+u^2right)^{frac{3}{2}}}spacetext{d}u=$$

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

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

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

Besides the other answers, I would like to expose another way of doing it:

$$int dfrac{x}{(a^2+x^2)^{3/2}} dx=-2int dfrac{1}{(a^2+x^2)^{1/4}}cdot dfrac {dfrac{-x}{2} (a^2+x^2)^{-3/4}}{(a^2+x^2)^{1/2}} , dx$$ 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.

You can take the solution as I've presented here:

begin{equation}
int_0^x frac{t^k}{left(t^n + aright)^m}:dt= frac{1}{n}a^{frac{k + 1}{n} - m} left[Bleft(m - frac{k + 1}{n}, frac{k + 1}{n}right) - Bleft(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^2right)^{frac{1 + 1}{2} - frac{3}{2}} left[Bleft(frac{3}{2} - frac{1 + 1}{2}, frac{1 + 1}{2}right) - Bleft(frac{3}{2} - frac{1 + 1}{2}, frac{1 + 1}{2}, frac{1}{1 + a^2x^2} right)right] \
&= frac{1}{left|aright|}left[Bleft(frac{1}{2}, 1right) - Bleft(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}
Bleft(frac{1}{2},1 right) = frac{Gammaleft(frac{1}{2} right)Gamma(1)}{Gammaleft(1 + frac{3}{2}right)} = frac{Gammaleft(frac{1}{2} right)Gamma(1)}{frac{1}{2}Gammaleft(frac{1}{2}right)} = 2Gamma(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|aright|}left[2 - Bleft(frac{1}{2},1, frac{1}{1 + a^2x^2} right) right]
end{equation}