Integrate $\int \frac{x^2}{(x^2+4)^2}$ $$\int \frac{x^2\;dx}{(x^2+4)^2}$$I suppose you must have to use trigonometric substitution, or something, but do not even know where to begin to solve this integral, guys, please help me!!
 A: Hint:  $$\frac{x^2}{(x^2+4)^2}=\frac1{x^2+4}-\frac4{(x^2+4)^2}$$
$$\therefore\int\frac{x^2}{(x^2+4)^2}dx=\int\frac1{x^2+4}dx-4\int\frac1{(x^2+4)^2}dx$$
A: Let 
\begin{align*}
x &= 2\tan \theta \\
dx &= 2 \sec^2 \theta \, d \theta \\
x^2 &= 4 \tan^2 \theta \\
x^2+4 &= 4 \tan^2 \theta +4 \\
&=4 \left( \tan^2 \theta +1 \right) \\
&=4 \sec^2 \theta.
\end{align*}
Then substitute,
\begin{align*}
\int \frac{x^2}{\left( x^2+4 \right)^2} \, dx &=\int \frac{4 \tan^2 \theta \cdot 2 \sec^2 \theta \, d \theta}{\left( 4 \sec^2 \theta \right)^2} \\
&= \int \frac{4 \tan^2 \theta \cdot 2 \sec^2 \theta}{16 \sec^4 \theta} \, d \theta \\
&=\frac{1}{2} \int \frac{\tan^2 \theta}{\sec^2 \theta} \, d \theta \\
&= \frac{1}{2} \int \sin^2 \theta \, d \theta \\
&= \frac{1}{4} \int \left( 1- \cos (2 \theta) \right) \, d \theta \\
&= \frac{1}{4} \left( \theta - \frac{1}{2} \sin (2 \theta) \right) \\
&=\frac{1}{4}\left( \theta -\cos \theta \cdot \sin \theta \right).
\end{align*}
For the back substitution, 
$$x=2 \tan \theta \Rightarrow \frac{x}{2} = \tan \theta \Rightarrow \theta = \tan^{-1}\left( \frac{x}{2}\right).$$
Form a right triangle with 2 adjacent to $\theta$, and $x$ opposite $\theta$. Then the hypoteneuse is $\sqrt{x^2+4}$. We use $\theta$, and read straight from the triangle to complete the back substitution.
\begin{align*}
\frac{1}{4}\left( \theta -\cos \theta \cdot \sin \theta \right) &= \frac{1}{4} \left( \tan^{-1} \left( \frac{x}{2} \right) - \frac{2}{\sqrt{x^2+4}} \cdot \frac{x}{\sqrt{x^2+4}} \right) \\
&=\frac{1}{4}\left( \tan^{-1}\left(\frac{x}{2}\right) -\frac{2x}{x^2+4} \right).
\end{align*}
A: Integrate by parts:
$$\int\frac{x^2}{(x^2+4)^2}dx=\int x\frac{x}{(x^2+4)^2}dx=\frac{1}{2} \int x(\frac{-1}{x^2+4})'dx=\frac{1}{2}(\frac{-x}{x^2+4} +\int \frac{1}{x^2+4}dx)=\frac{1}{2}(\frac{-x}{x^2+4} +\frac{1}{2}\arctan\frac{x}{2}) +C$$
