Thread: Help with an integral

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

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
$\displaystyle
(1+x^4)=(1-\sqrt 2 x+x^2)(1+\sqrt 2 x+x^2)$

Originally Posted by thelostchild

Evaluate the following integral, without using partial fractions

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

first divide by $\displaystyle x^2$

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

Add zero to the denominator by adding and taking away 2

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

Compete "the square"

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

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

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

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

(skipping some steps) ....

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


$\displaystyle \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