How do I integrate $\int \frac{x^2}{(x^2+9)^2}\, dx$?

Question

How do I integrate $\displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx$ ?

I tried doing using algebra and solving the question a bit. But it didn't become something that looks solvable.

How should I do this?

Answer 1

I think the best thing to do in such a question, should be suitable substitution. So let

$x=3\tan{\theta}$

$dx=3\sec^2(\theta) d\theta$

$\begin{align} \displaystyle \int \frac{x^2}{(x^2+9)^2}\, dx & = \int \frac{9 \tan^2(\theta) }{(9 \tan^2(\theta) + 9)^2} 3\sec^2(\theta)\, d\theta = \int \frac{27 \tan^2(\theta)\sec^2(\theta)}{81 \sec^4(\theta)} \, d\theta \\ & = \int \frac{1}{3} \frac{ \tan^2(\theta)}{ \sec^2(\theta)}\, d\theta = \frac{1}{3} \int \sin^2(\theta) \, d\theta \\ & = \frac{1}{3} \int \frac{1}{2} - \frac{\cos(2\theta)}{2} \, d\theta = \frac{1}{3} (\frac{1}{2}\theta - \frac{\sin(2\theta)}{4}) + C\\ & = \frac{1}{6} \theta - \frac{2\sin\theta\cos\theta}{12} + C = \frac{1}{6} \theta - \frac{1}{6} \sin{\theta}\cos{\theta} + C \end{align}$

And since $x=3\tan \theta$, we get that:

• $\displaystyle \theta = \arctan(\frac{x}{3})$
• $\displaystyle \cos \theta = \frac{3}{\sqrt{x^2+9}}$
• $\displaystyle \sin \theta = \frac{x}{\sqrt{x^2+9}}$ Therefore, turning $\theta$ back into $x$, we get that: $\displaystyle \begin{align}\int \frac{x^2}{(x^2+9)^2} \, dx & = \frac{1}{6} \theta - \frac{1}{6} \sin{\theta}\cos{\theta} + C \\ & = \frac{1}{6} \arctan(\frac{x}{3}) - \frac{1}{6} \frac{x}{\sqrt{x^2+9}}\frac{3}{\sqrt{x^2+9}} + C \\ & = \frac{1}{6} \arctan(\frac{x}{3}) - \frac{1}{2} \left(\frac{x}{x^2+9}\right)+C\end{align}$

And that should be your final answer.

Answer 2

Hint:

Integrate by parts

$$\int uv\ dx=u\int v\ dx-\int\left(\dfrac{du}{dx}\int v\ dx\right)dx$$

Here $u=x,v=\dfrac x{(9+x^2)^2} $

Answer 3

Hint:

Use $x=3\tan t,\implies dx=?$ and $x^2+9=9\sec^2t$

finally use $\cos2y=1-2\sin^2y$