Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.

Hint:
$4x^4+25$ factors as the product of two irreducible quadratic polynomials:
$$4x^4+25=(2x^2+5)^2-20x^2=(2x^2-2\sqrt 5x+5)(2x^2+2\sqrt 5x+5,$$
whence the partial fractions decomposition:
$$\frac{x^2}{4x^4+25}=\frac{Ax+B}{2x^2-2\sqrt 5x+5}+\frac{Cx+D}{2x^2+2\sqrt 5x+5}.$$
As $\dfrac{x^2}{4x^4+25}$ is an even function, we have $C=-A,\enspace B=D$. Setting $x=0$ shows $B=D=0$, and reducing to the same denominator yields $A=\dfrac{\sqrt 5}{20}$. Hence we have to compute:
$$\frac{\sqrt5}{20}\int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x-\frac{\sqrt5}{20}\int\frac{x}{2x^2+2\sqrt 5x+5}\,\mathrm d x$$
Now split each integral:
\begin{align*}
\int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x&= \frac14\int\frac{4x-2\sqrt5}{2x^2-2\sqrt 5x+5}\,\mathrm d x+\frac12\int\frac{\mathrm d x}{4x^2-4\sqrt 5x+10} \\[1ex]
&=\frac14\ln(2x^2-2\sqrt 5x+5)+\frac14\int\frac{\mathrm d(2x-\sqrt5)}{(2x-\sqrt5)^2+5} \\[1ex]
&= \frac14\ln(2x^2-2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x-\sqrt5)
\end{align*}
Similarly,
$$\int\frac{ x}{2x^2+2\sqrt 5x+5}\,\mathrm d x=\frac14\ln(2x^2+2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x+\sqrt5)$$
Thus we obtain
$$\int\frac{x^2}{4x^4+25}\,\mathrm dx=\frac{\sqrt5}{80}\ln\biggl(\frac{2x^2-2\sqrt 5x+5}{2x^2+2\sqrt 5x+5}\biggr)+\frac{\sqrt5}{400}(\arctan(2x-\sqrt5)-\arctan(2x+\sqrt5)). $$
The last term may be further simplified using the formula $\;\arctan p-\arctan q \equiv\arctan\dfrac{p-q}{1+pq}\mod\pi$.
