# Thread: Help with an integral

1. ## Help with an integral

Hey I'm having a little trouble with this integral
$
\int \frac{1+x^2}{1+x^4}dx$

I know the answer just need a little nudge in the right direction with it

many thanks

Simon

EDIT dont worry I got it, stupid me forgetting
$
(1+x^4)=(1-\sqrt 2 x+x^2)(1+\sqrt 2 x+x^2)$

2. Originally Posted by thelostchild
Evaluate the following integral, without using partial fractions

$\int \frac{1+x^2}{1+x^4}dx$

first divide by $x^2$

$\Rightarrow \int \frac{1+x^{-2}}{x^{-2}+x^{2}}dx$

Add zero to the denominator by adding and taking away 2

$\Rightarrow \int \frac{1+x^{-2}}{ x^{2} - 2 + x^{-2} + 2}dx$

Compete "the square"

$\Rightarrow \int \frac{1+x^{-2}}{(x - x^{-1})^2 + 2}dx$

Now use the subistution $u = x - x^{-1} \ \ \ \Rightarrow \ \ \ du = (1+x^{-2})dx$

that makes the integral. $\int \frac{1}{u^2 + 2}du$

use another subistution $u = \sqrt{2} \tan \theta$

(skipping some steps) ....

$\int \frac{1}{u^2 + 2}du = \frac{1}{ \sqrt{2}}\arctan{\frac{u}{ \sqrt{2}}} + C$

$\therefore \int \frac{1+x^2}{1+x^4}dx = \frac{1}{ \sqrt{2}}\arctan \left( \frac{ x - x^{-1}}{ \sqrt{2}} \right) + C$

Bobak
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