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I would appreciate any hints on how to solve the following integration problems, they are my homework questions btw:

$$\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$$

$$\int \frac{x^4}{x^4+5x^2+4} \, dx$$

Thank you very much in advance!

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Hint Have you tried dividing the fractions and then using partial fractions, if necessary? – Daryl Nov 13 '12 at 3:26
up vote 1 down vote accepted

Here are a couple of off-hand suggestions. There are undoubtedly more efficient ways.

For the first integral, you could multiply out the top, and use polynomial long division. You will get something that has the shape $P(x)+\frac{Ax}{1+x^2}+\frac{B}{1+x^2}$.

Integrating the polynomial will be easy. For $\frac{Ax}{1+x^2}$, use $u=1+x^2$.

For the second problem, divide. We get $1-\frac{5x^2+4}{x^4+5x^2+4}$. Then use the fact that $x^4+5x^2+4=(x^2+1)(x^2+4)$, and use partial fractions. We end up needing to integrate $\frac{A}{x^2+1}$ and $\frac{B}{x^2+4}$. The first is immediate. For the second, use $x=2u$.

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Why the partial fractions cannot be $\frac{Ax+B}{x^2+1} $ and $\frac{Bx+C}{x^2+1} $? – uohzxela Nov 13 '12 at 4:01
In the second problem? Because the top $5x^2+4$ is symmetric about $0$. But anyway, do the partial fraction decomposition explicitly. You will see that there are no "$x$" terms. (If there were such terms, they would not be a big problem, but there happen not to be any.) – André Nicolas Nov 13 '12 at 4:19

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