# Can't compute this integral

I would appreciate your help solving this integral:

$$\int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x$$

I tried using linear substitution with $t = x/a$ and then trying to bring it to some combination of the known integral of $\arctan (x) = \frac{1}{x^2+1}$ but I'm not sure it will be helpful because there isn't just $1$ in the numerator.

Basically, I got stuck very early in the process:

$$\int_2^3 \frac{1+x^3}{(a^2(\frac{x^2}{a^2}+1))^\frac{3}{2}}\mathrm{d}x$$

Thank you.

• Ah I see your confusion, pretty hard to compute an integral without a differential. – electronpusher May 6 '17 at 21:24
• The antiderivative of $$\frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}$$ is $$\frac{2a^4+a^2x^2+x}{a^2\sqrt{a^2+x^2}}+C.$$ – zoli May 6 '17 at 21:38
• @zoli, I got $frac{x+a^2\sqrt{x^2+a^}+2a^4}{a^2\sqrt{x^2+a^2}}$ – electronpusher May 6 '17 at 21:43
• wolframalpha.com/input/… – zoli May 6 '17 at 21:45

I suggest breaking it into two integrals, and using the trig substitution $x=a\tan t$ to turn each one into a trigonometric integral. Thus:

\begin{align} \int_2^3 \frac{1+x^3}{(x^2+a^2)^\frac{3}{2}}\mathrm{d}x &= \int_2^3 \frac{dx}{(x^2+a^2)^{3/2}} + \int_2^3\frac{x^3}{(x^2+a^2)^{3/2}}dx\\ &=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sec^2 t}{a^3\sec^3 t}dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a^3 \tan^3 t \cdot a\sec^2 t}{a^3\sec^3 t}dt\\ &=\int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{1}{a^2}\cos t dt + \int_{\tan^{-1}(2/a)}^{\tan^{-1}(3/a)}\frac{a\sin^3 t}{\cos^2 t}dt \end{align}

Can you get it from there?

• Thank you for commenting but no so sure I followed - what is the meaning of $\sec$? – Noam May 6 '17 at 22:29
• $\sec$ is the trigonometric function secant, the reciprocal of cosine. – G Tony Jacobs May 7 '17 at 0:02

Hint: Let $x \mapsto a \tan \theta$, then $\mathrm{d}x = a(1 + \tan \theta) \,\mathrm{d} \theta$. Then

$$\int \frac{1+x^3}{(a^2+x^2)^{3/2}} = \frac{1}{a^2}\int \frac{\sin^3\theta}{\cos ^2\theta}+\cos \theta\,\mathrm{d}\theta$$

For the first integral use $\sin^3 \theta = \sin \theta(1-\cos^2\theta)$ and set $y \mapsto \cos \theta$. The latter integral is trivial.

Your integral can be split into $$\int\frac{dx}{(x^2+a^2)^{3/2}}+\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}}$$

The first integral can be solved by trig substitution. U-substitution can be used on the second integral to obtain

$$\int\frac{x^2\cdot xdx}{(x^2+a^2)^{3/2}} =\frac{1}{2}\int\frac{u-a^2}{u^{3/2}}du \\ = \frac{1}{2}\int(u^{-1/2}-a^2u^{-3/2})du$$.

And it's busywork from there.

• thank you for commenting but I think you used $x^2$ instead of $x^3$? and can you elaborate on the trig substitution you made? – Noam May 6 '17 at 22:31
• I used the same trig sub explained in G Tony Jacobs' answer; however, I did not apply it to the second term. For the second term I used the substitution $$u=x^2+a^2$$. I separated $$x^3=x^2 \cdot x$$ to obtain $$xdx=1/2du$$, and the leftover $$x^2=u-a^2$$ (according to the substitution I defined). I recommend trying to go through the steps yourself to fully understand. – electronpusher May 7 '17 at 0:31