Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$

Question

Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$

My Approach:

$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$

But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possible which makes sense since you can't take a small element of the form $(x^2+a^2)$. How can I then solve this integration without using trigonometry?

Doing $u$-substitution here, which is what you tried, is possible but $d(x^2+a^2)=2xdx$ introduces an $x$ that you'll have to turn into a square root and you won't have gained anything. — W. Cadegan-Schlieper, Apr 07 '17 at 15:04
What you're trying to do is rewrite the integral in terms of $u=a^2+x^2$, but the converted integral will introduce a factor of $x^{-1}=(u-a^2)^{-1/2}$. — W. Cadegan-Schlieper, Apr 07 '17 at 15:09
@W.Schlieper wouldn't we simply solve the integration to get $\frac{-1}{2(2x)} (a^2+x^2)^{-1/2}$ — oshhh, Apr 07 '17 at 15:13
@OsheenSachdev Oftentimes, you can use this automatic tool for evaluating integrals. Please try to use this tool before asking a question. (note: This is not spam, and I am not affiliated with the site; posted only to help the OP and future users) — Pritt Balagopal, Jul 13 '17 at 08:41

Mark Pineau · Answer 1 · 2017-07-13T08:37:06.180

Here is the correct solution. Note for anything of the form $({x^2+a^2)^{n/2}}$ for n odd, consider using the following trigonometric substituion.

So, let $x=atan(u) \ (or \ x=asinh(u)$ works aswell$)$ $\Rightarrow dx=(a)sec^2(u)du$

Thus we have:

$$\int \frac{1}{(x^2+a^2)^{3/2}}dx=\int \frac {asec^2u}{(a^2tan^2(u)+a^2)^{3/2}}du=\frac {a}{a^3}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du$$

$$=\frac {1}{a^2}\int \frac {sec^2u}{(tan^2(u)+1)^{3/2}}du=\frac {1}{a^2}\int \frac {sec^2u}{(sec^2(u))^{3/2}}du=\frac {1}{a^2}\int \frac {1}{sec(u)}du=\frac {1}{a^2}\int cos(u)du$$

Finally, we have:

$$=\frac{1}{a^2}sin(u)+c=\frac{1}{a^2}sin(arctan(\frac xa))+c=\frac{x}{a^2 \sqrt{a^2+x^2}}+c$$

Bérénice · Accepted Answer · 2017-04-07T15:48:25.020

4

Following your idea we get $du=2xdx$ so : $$I=\int (x^2+a^2)^{-3/2} dx=\int\frac{2xdx}{2x(x^2+a^2)^{3/2}}=\int \frac{du}{2\sqrt{u^4-a^2u^3}}=\frac{1}{2}\int\frac{du}{u^{3/2}\sqrt{u-a^2}} $$ Let $v=\frac{1}{u-a^2}$, we get $dv=-\frac{1}{(u-a^2)^2}$ : $$I=-\frac{1}{2}{\displaystyle\int}\dfrac{1}{\left(a^2v+1\right)^\frac{3}{2}}\,\mathrm{d}v$$ Let $w=a^2v+1$ then $dw=a^2$ : $$I=-\frac{1}{2}\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{a^2}}}{\displaystyle\int}\dfrac{1}{w^\frac{3}{2}}\,\mathrm{d}w=\dfrac{1}{a^2\sqrt{w}}$$

And finally : $$I=\dfrac{1}{a^2\sqrt{\frac{x}{x-a^2}}}+C$$

edited Apr 07 '17 at 15:48

answered Apr 07 '17 at 15:14

Bérénice

9,487

Oh god...this is so long! – oshhh Apr 07 '17 at 15:19
Could you also help me figure out the error in the answer I wrote in the comment by simply integrating the expression I formulated and then dividing by the derivative $d(x^2+a^2)/dx$ – oshhh Apr 07 '17 at 15:21
It is faster you can do the trigonometric substitution $u=\arctan(x/a)$ – Bérénice Apr 07 '17 at 15:22
1

Hmm when youmake a subsitution $u=f(x)$ you obtain an expression of $du$ in term of $dx$ and you want to make appear this expression inside your integral, in order to change this expression into $du$ after teh change of variable. You can seefor example in the first line I multiplied both sides of the fraction by $2x$ in order to make appear $du$ and to be able to pursue the change of variable. – Bérénice Apr 07 '17 at 15:29
Got it...thank you :) – oshhh Apr 07 '17 at 15:32
This is correct, the substitution is clever, but when in doubt, refer to the trigonometric substitutions. There are rules regarding when to substitute what, simplifying the integral much faster for most questions of these type. – Mark Pineau Apr 07 '17 at 15:40
@GandalftheMathWizard But OP did not want to use a trig substitution :(. But I agree it is much faster :) – Bérénice Apr 07 '17 at 15:46
I found a usefull edit :3 – Bérénice Apr 07 '17 at 15:48
Ight all good, didn't want to down vote a correct solution, cheers! – Mark Pineau Apr 07 '17 at 15:49

score 2 · Answer 3 · answered Apr 07 '17 at 15:20

We use a trick to get a reduction formula: $$ \frac{1}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x^2+a^2-x^2}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{1}{\sqrt{x^2+a^2}} - \frac{1}{a^2}\frac{x^2}{(x^2+a^2)^{3/2}}. $$ The first term is lower-order, the second can be integrated by parts: $$ \int \frac{x^2}{(x^2+a^2)^{3/2}} \, dx = -\frac{x}{\sqrt{x^2+a^2}} + \int \frac{dx}{\sqrt{x^2+a^2}}. $$ Oh, but now the new integral on the right here cancels with the integral of the first fraction above! Hence $$ \int \frac{dx}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x}{\sqrt{x^2+a^2}} +C, $$ and it's easy to verify this by differentiating.

Integrate $x/(x^2+a^2)^{3/2}$ (it's an $f'(x)f(x)^a$ form), differentiate the other $x$. — Chappers, Apr 07 '17 at 15:25

score 0 · Answer 4 · answered Feb 11 '25 at 10:45

What about:

$$ \int \left(x^2 + a^2\right)^{-\frac{3}{2}} dx$$

$$ = \int \left(x^2 \left(1 + a^2x^{-2}\right)\right)^{-\frac{3}{2}} dx$$

$$ = \int x^{-3}\left(1 + a^2x^{-2}\right)^{-\frac{3}{2}} dx$$

If you set $u = 1 + a^2x^{-2}$, then you can complete it from here. I hope this helps! :) (I know this answer is nearly 8 years late but hopefully it's useful to anyone else reading this, because I think it's the simplest non-trig approach.)

Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$

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