# partial fractions

• Feb 22nd 2009, 04:47 PM
razorfever
partial fractions
find the antiderivative of:

(x^3 - 5x^2 - 4) / (x^4 - 4x^2)

i used the difference of two squares to get it into this form

(x^3 - 5x^2 - 4) = (Ax + B)(x^2 - 2x) + (Cx + D)(x^2 - 2x)

but i can't seem to solve for A,B,C,D and I'm not getting any term equal to the -4 on the left side, what am i doing wrong

can someone post a step by step solution from this point onwards?
• Feb 22nd 2009, 04:51 PM
TheEmptySet
Quote:

Originally Posted by razorfever
find the antiderivative of:

(x^3 - 5x^2 - 4) / (x^4 - 4x^2)

i used the difference of two squares to get it into this form

(x^3 - 5x^2 - 4) = (Ax + B)(x^2 - 2x) + (Cx + D)(x^2 - 2x)

but i can't seem to solve for A,B,C,D and I'm not getting any term equal to the -4 on the left side, what am i doing wrong

can someone post a step by step solution from this point onwards?

try this factorization for the denominator

$\displaystyle (x^4-4x^2)=x^2(x^2-4)=x^2(x-2)(x+2)$
• Feb 22nd 2009, 05:08 PM
razorfever
oh nice
but why doesn't the one i'm doing work?
what's wrong with it?
it seems like the more obvious way to go right?
• Feb 22nd 2009, 05:24 PM
Krizalid
$\displaystyle \int{\frac{x^{3}-5x^{2}-4}{x^{4}-4x^{2}}\,dx}=\underbrace{\int{\frac{x}{x^{2}-4}\,dx}}_{\alpha }\,-\,5\underbrace{\int{\frac{dx}{x^{2}-4}}}_{\beta }\,-\,4\underbrace{\int{\frac{dx}{x^{4}-4x^{2}}}}_{\gamma }.$

$\displaystyle \alpha$ is easy to solve, so I'll take care of $\displaystyle \gamma.$ Put $\displaystyle x=\frac1u$ and the integral is

\displaystyle \begin{aligned} \int{\frac{dx}{x^{4}-4x^{2}}}&=\int{\frac{u^{2}}{4u^{2}-1}\,du} \\ & =\frac{1}{4}\int{\frac{4u^{2}-1+1}{4u^{2}-1}\,du} \\ & =\frac{1}{4}\left\{ \int{du}+\underbrace{\int{\frac{du}{(2u+1)(2u-1)}}}_{\varphi } \right\}. \end{aligned}

Now I'll solve $\displaystyle \beta,$ thus we have $\displaystyle \int{\frac{dx}{x^{2}-4}}=\frac{1}{8}\int{\frac{(x+4)-(x-4)}{(x+4)(x-4)}\,dx}=\frac{1}{8}\left( \int{\frac{dx}{x-4}}-\int{\frac{dx}{x+4}} \right).$ Finally, by using this procedure, you can easily solve $\displaystyle \varphi,$ thus, no more integrals and we're done.