# How do you find the integral of int dx / (x²+4)²?

Oct 23, 2015

I gave the wrong answer. Sorry!!! It has been too long since I did any Calculus. The correct answer is:$\frac{x}{8 \left({x}^{2} + 4\right)} + \frac{1}{16} \arctan \left(\frac{x}{2}\right)$

My original post deleted

#### Explanation:

Solution checked in Maple. I need to do a self study refresher on my calculus before I answer any more of that type of question. Sorry again!!!

Oct 23, 2015

I found: $\frac{1}{16} \left\{\arctan \left(\frac{x}{2}\right) + \frac{1}{2} \sin \left\{2 \arctan \left(\frac{x}{2}\right)\right]\right\} + c$ but PLEASE check my maths!!!!

#### Explanation:

I would try to manipulate the denominator setting
$x = 2 t$ so that $\mathrm{dx} = 2 \mathrm{dt}$:

$\int \frac{2 \mathrm{dt}}{4 {t}^{2} + 4} ^ 2 = \int \frac{2 \mathrm{dt}}{{4}^{2} {\left({t}^{2} + 1\right)}^{2}} = \frac{1}{8} \int \frac{\mathrm{dt}}{{t}^{2} + 1} ^ 2 =$

now let us set $t = \tan \left(u\right)$ so that $\mathrm{dt} = \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du}$
and:

$= \frac{1}{8} \int \frac{1}{{\tan}^{2} \left(u\right) + 1} ^ 2 \cdot \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du} =$
$= \frac{1}{8} \int \frac{1}{\left(\frac{{\sin}^{2} \left(u\right) + {\cos}^{2} \left(u\right)}{\cos} ^ 2 \left(u\right)\right)} ^ 2 \cdot \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du} =$
$= \frac{1}{8} \int \frac{{\cos}^{4} \left(u\right)}{1} \cdot \frac{1}{\cos} ^ 2 \left(u\right) \mathrm{du} = = \frac{1}{8} \int {\cos}^{2} \left(u\right) \mathrm{du} =$

here we can use integration by parts and solve it as (if you cannot tell me, I'll write it for you, I do not want to make it too long here):

$= \frac{1}{8} \cdot \frac{1}{2} \left(u + \frac{1}{2} \sin \left(2 u\right)\right) + c =$

but $u = \arctan \left(t\right)$ so:

$= \frac{1}{16} \left\{\arctan \left(t\right) + \frac{1}{2} \sin \left[2 \arctan \left(t\right)\right]\right\} + c =$

and $t = \frac{x}{2}$ so finally...

$= \frac{1}{16} \left\{\arctan \left(\frac{x}{2}\right) + \frac{1}{2} \sin \left\{2 \arctan \left(\frac{x}{2}\right)\right]\right\} + c$