# Thread: looks simple Integral but it is fake

1. ## looks simple Integral but it is fake

Hi . Can anyone give me a way to solve this Integral :

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

Regards ...

2. ## Try

Try this...

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

3. EDIT:: Made a mistake. Re-checking everything...

Sub x^2 = tan theta.
x = sqrt(tan theta)
$\displaystyle dx = \frac12 (tan \theta)^{-\frac12}.sec^2\theta$

$\displaystyle \int \frac{x^2}{1+x^4} dx = \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta$

$\displaystyle \int \frac{tan\theta}{1+tan^2 \theta} . \frac12 (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta$

$\displaystyle \frac12 \int \frac{tan\theta}{sec^2\theta} . (tan \theta)^{-\frac12}.sec^2\theta d\theta= \frac12 \int tan\theta(tan \theta)^{-\frac12} d\theta = \frac12 \int (tan\theta)^{\frac12} d\theta$

$\displaystyle \frac12 \int (tan\theta)^{\frac12} d\theta = \frac12 [\frac{2(tan\theta)^{\frac32}}{3sec^2\theta}]$

Not sure if that's possible...

4. Originally Posted by Unknown008
You can use a trigonometric substitution...

let x^4 = tan^2(\theta).

$\displaystyle \int \frac{x^2}{1+x^4} dx = \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}} d\theta$

$\displaystyle \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}}d \theta= \int \frac{tan(\theta)}{\sqrt{sec^2(\theta)}}d\theta = \int \frac{tan(\theta)}{sec(\theta)} d\theta= \int sin(\theta)d\theta = -cos(\theta)$

$\displaystyle -cos(\theta) = -\frac{1}{\sqrt{x^2+1}}$
How did you get a square root in the den?

$\displaystyle \int \frac{x^2}{1+x^4} dx = \int \frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}} d\theta$

What did you sub dx with?

5. Oops, sorry, I made a big mistake... I messed things up... >.<

Ok, I'll be looking for something else...

6. Originally Posted by Unknown008
Oops, sorry, I made a big mistake... I messed things up... >.<

Ok, I'll be looking for something else...
Lol it ok, I just kept trying and could not do it the way you tried. Thought I was going crazy lol. I tried subbing in a hyperbolic trig function and that did not work. I solved it using a power series but I'm not sure if that is what you are looking for?

7. Originally Posted by Also sprach Zarathustra
Try this...

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

If the second integral has negative 1 in the numerator then

$\displaystyle x^2 - 1 - 1$ DNE $\displaystyle x^2$

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

$\displaystyle =\int \left( \sum_{n=0}^{\infty} (-1)^n x^{4n+2} \right) \, dx$

$\displaystyle =C+\sum_{n=0}^{\infty} (-1)^n \frac{x^{4n+3}}{4n+3}$