# How do you solve this integral?

• Sep 24th 2008, 12:48 PM
cityismine
How do you solve this integral?
integral [ (x^3+4x^2-x+2)/(2x^4+5x^2+2) dx]

I tried solving it by using partial fractions, but answer my answer doesn't match the one in the back of the book.

I put it in the following format and attempted to solve for the constants.

(Ax+B)/(2x^2+1)+(Cx+D)/(x^2+2)
• Sep 24th 2008, 12:59 PM
Moo
Hello,
Quote:

Originally Posted by cityismine
integral [ (x^3+4x^2-x+2)/(2x^4+5x^2+2) dx]

I tried solving it by using partial fractions, but answer my answer doesn't match the one in the back of the book.

I put it in the following format and attempted to solve for the constants.

(Ax+B)/(2x^2+1)+(Cx+D)/(x^2+2)

Can you show your working so that we can check together what's wrong ?

If you want to check, the partial fractions should be :

$\frac{x+2}{x^2+2}-\frac{x}{2x^2+1}$
• Sep 24th 2008, 01:22 PM
cityismine
Those are the exact fractions I got, I probably made a mistake integrating afterwards. Let me check my work...
• Sep 24th 2008, 02:31 PM
cityismine
How do you integrate (x+2)/(x^2+2). I believe this is the step, that I'm having trouble with.
• Sep 24th 2008, 02:46 PM
javax
Quote:

Originally Posted by cityismine
How do you integrate (x+2)/(x^2+2). I believe this is the step, that I'm having trouble with.

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

now for the first integral just substitute $t=x^2+2$ and you're done, as for the second it's a tabular one!