Did I solve this integral correctly? (trig substitution) I'm having trouble with trig substitution. This is what I've done so far, but I'm not sure if I did everything right. This is the integral:
$$\int \frac{x^2}{(1+x^2)^\frac{3}{2}}$$
and my substitution is:
$x= \tan\Theta$
$$\int \frac{\tan\Theta ^2}{(\sqrt{\sec\Theta ^2})^3}\sec\Theta ^2d\Theta $$
$$\int \frac{\sec\Theta ^2-1}{\sec\Theta }$$
$$\int \sec\Theta -\int \cos\Theta $$
$$(\ln(\sec\Theta +\tan\Theta ))-(\sin\Theta)$$
$$\ln(\frac{1}{\sqrt{1-x^2}}+ \frac{x}{\sqrt{1-x^2}}) - x + C$$
Thanks!

Okay up to the last step.
Now 
$$
\tan(\theta) = x \\
\sec(\theta) = \sqrt{1+x^2}\\
\sin(\theta) = {\frac{x}{\sqrt{x^2+1}}} $$
So your answer is
$$\log \left(\sqrt{x^2+1}+x\right)-{{x}\over{\sqrt{x^2+1}}}+C$$
Differentiate the answer and square it to get
$${\frac{x^4}{\left(x^2+1\right)^3}}$$
