1. ## 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?

2. 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

$(x^4-4x^2)=x^2(x^2-4)=x^2(x-2)(x+2)$

3. 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?

4. $\int{\frac{x^{3}-5x^{2}-4}{x^{4}-4x^{2}}\,dx}=\underbrace{\int{\frac{x}{x^{2}-4}\,dx}}_{\alpha }\,-\,5\underbrace{\int{\frac{dx}{x^{2}-4}}}_{\beta }\,-\,4\underbrace{\int{\frac{dx}{x^{4}-4x^{2}}}}_{\gamma }.$

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

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