Did I solve this integral correctly? (trig substitution)

Question

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!

Answer 1

The problem here is simply when "back substituting".

Having set $x =\tan \theta \implies (\sec\theta = \sqrt{1 + x^2} \,\text{ and }\, \sin \theta = \dfrac{x}{\sqrt{1+x^2}})$

This gives us the final answer: $$\ln \left(\sqrt{x^2+1}+x\right)-{{x}\over{\sqrt{x^2+1}}}+C$$

Answer 2

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 $${\frac{x^4}{\left(x^2+1\right)^3}}$$