# Integration questions on $\int \frac{x^4\left ( 1-x \right )^4}{1+x^2} \, dx$ and $\int \frac{x^4}{x^4+5x^2+4} \, dx$

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

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 $x=2u$.
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