Integration by partial fractions?

**Fausst** · October 21st 2009, 08:23 PM

how do you solve this?:

$ \int \frac {x^3-6}{x^4+6x^2+8} dx $

**calum** · October 21st 2009, 08:35 PM

The first step you'll need to do is to factorise the denominator. Judging by that polynomial, you will end up with two irreducible quadratic factors. I'll give you a hint on a quick way of factorising: there is a 'hidden quadratic'. Substitute a for $x^2$ and get two factors in terms of a, then convert back to factors in terms of x

**pickslides** · October 21st 2009, 08:36 PM

Consider

$\frac {x^3-6}{x^4+6x^2+8} = \frac {x^3-6}{(x^2+4)(x^2+2)}$

**redsoxfan325** · October 21st 2009, 08:46 PM

Originally Posted by Fausst

how do you solve this?:

$ \int \frac {x^3-6}{x^4+6x^2+8} dx $

Since $x^4+6x^2+8=(x^2+2)(x^2+4)$ , that fraction decomposes to:

$\frac{x^3-6}{x^4+6x^2+8}=\frac{Ax+B}{x^2+4}+\frac{Cx+D}{x^2+ 2}$

Solve for $A,B,C,D$ and then integrate the two pieces separately. To do this, break up $\int\frac{Ax+B}{x^2+4}\,dx$ (for whatever values of $A$ and $B$ you get) into $\frac{A}{2}\int\frac{2x}{x^2+4}\,dx+\int\frac{B}{x ^2+4}\,dx=$ $\frac{A}{2}\ln(x^2+4)+\frac{1}{4}\int\frac{B}{(x/2)^2+1}\,dx$

The latter integral will integrate to something involving $\arctan$ , but I leave the actual calculations to you.

**calum** · October 21st 2009, 08:51 PM

Looking at this again, the first thing you should do is some substitution to get that cubic out of the numerator.

Let $a = x^4 + 6x^2 + 8$ then $\frac{da}{dx} = 4x^3 + 12x$ so $dx = \frac{da}{4(x^3 + 3x)}$

Now you can make the integral be $\int \frac{x^3 + 3x}{4(x^3 + 3x)a} da + \int \frac{-3x - 6}{x^4 + 6x^2 + 8}dx$

**redsoxfan325** · October 21st 2009, 08:54 PM

Originally Posted by calum

Looking at this again, the first thing you should do is some substitution to get that cubic out of the numerator.

Let $a = x^4 + 6x^2 + 8$ then $\frac{da}{dx} = 4x^3 + 12x$ so $dx = \frac{da}{4(x^3 + 3x)}$

Now you can make the integral be $\int \frac{x^3 + 3x}{4(x^3 + 3x)a} da + \int \frac{-3x - 6}{x^4 + 6x^2 + 8}dx$

That's true, but he still needs to do partial fraction decomposition on the second integral, so this substitution might not save much (if any) time.

**Fausst** · October 21st 2009, 09:23 PM

Thanks for replying

Originally Posted by redsoxfan325

Solve for $A,B,C,D$

I found that: A=2 B=3 C=-1 D=-3

but i don't understand what you did to separate

$\int\frac{Ax+B}{x^2+4}\,dx $

**redsoxfan325** · October 21st 2009, 09:44 PM

Originally Posted by Fausst

Thanks for replying

I found that: A=2 B=3 C=-1 D=-3

but i don't understand what you did to separate

$\int\frac{Ax+B}{x^2+4}\,dx $

Remember that the integral of a sum is the sum of the integrals. So

$\int\frac{Ax+B}{x^2+4}\,dx=\int\frac{Ax}{x^2+4}\,d x+\int\frac{B}{x^2+4}\,dx$

$\int\frac{Ax}{x^2+4}\,dx=\int\frac{\frac{A}{2}\cdo t2x}{x^2+4}\,dx=\frac{A}{2}\int\frac{2x}{x^2+4}\,d x$

because you can pull constants out in front of integrals.

**redsoxfan325** · October 22nd 2009, 09:03 AM

@ Fausst: I had typo in my original post, which I have fixed.

Also, the general rule for $\arctan$ integration is $\int\frac{c}{x^2+a}\,dx=\frac{c\sqrt{a}}{a}\tan^{-1}\left(\frac{x}{\sqrt{a}}\right)$ , where $a$ and $c$ are arbitrary constants.

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