1. ## Basic Intergration Question

Looking to integrate the following.

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

We usually have to re-write the denominator as a negative and solve that way yeah? For example...

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

is

$\displaystyle \int x^2 + x^-2$ and integrate.

How do I solve it though with the +1 on the denominator? This might be a really really stupid question, but I can't simplify..... can I?

P.s. This is more of a problem about simplifying rather than integration I suspect.

2. I'm not that great at calculus, but I noticed your problem and think I may know the answer. The key is in the following:

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

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

You should be right integrating that simplified form, but just in case you need the answer for reference, it's:

$\displaystyle \frac {x^3}{3} - x + arctan(x)$

3. Originally Posted by drew.walker
I'm not that great at calculus, but I noticed your problem and think I may know the answer. The key is in the following:

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

$\displaystyle \int {\frac {(x^2 + 1)(x^2 - 1) + 1} {x^2 + 1}} dx = \int {x^2 - 1 + \frac {1}{x^2 + 1}} dx$
$\displaystyle \frac {x^3}{3} - x + arctan(x)$