How to simplify the integration? If any integration is in form $$\int \frac{1}{1+x^2}dx$$ it easily follows the $\tan^{-1}x$ but How to simplify if we have
$$\int \frac{1}{(1+x^2)^2}dx$$
 A: Since 
$$\left(\frac{x}{1+x^2}\right)'=\frac{2-(1+x^2)}{(1+x^2)^2}=\frac{2}{(1+x^2)^2}-\frac{1}{1+x^2}$$
one has
$$2\int \frac{1}{(1+x^2)^2}dx=\frac{x}{1+x^2}+\int \frac{1}{1+x^2}dx$$
$$\Rightarrow \int\frac{1}{(1+x^2)^2}dx=\frac{x}{2(1+x^2)}+\frac{\tan^{-1} x}{2}+C$$
A: by using partial fraction 
$$\frac{1}{(1+x^2)^2}=\frac{1}{2}(\frac{1}{1+x^2}-\frac{x^2-1}{(x^2+1)^2})$$
$$=\frac{1}{2}(\frac{1}{1+x^2}-\frac{d}{dx}(\frac{x}{(x^2+1)}))$$
A: Define $f(\alpha)=\int\dfrac{dx}{\alpha+x^2}=\frac{1}{\sqrt\alpha}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)$ 
You can recover the original integral by putting $\alpha=1$ later, 
Now, differentiating both sides.
$f'(\alpha)=\int\dfrac{-dx}{(\alpha+x^2)^2}=\frac{-1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{-x}{2(\alpha)^{3/2}}$ 
$\int\dfrac{dx}{(\alpha+x^2)^2}=\frac{1}{2(\alpha)^{3/2}}\tan^{-1}\left(\frac{x}{\sqrt\alpha}\right)+\frac{1}{(\alpha)^{1/2}}\frac{1}{1+\frac{x^2}{\alpha}}\frac{x}{2(\alpha)^{3/2}}$
Put $\alpha=1$
$$\int\dfrac{dx}{(1+x^2)^2}=\frac{1}{2}\tan^{-1}x+\frac{1}{1+x^2}.\frac{x}{2}$$ and of course the remaining $C$
And there you have your integral found by differentiating only.^^
A: An idea that works for $\frac{1}{(1+x^2)^2}$ and for all other exponents $\frac{1}{(1+x^2)^n}$ is to get the base of the denominator in the numerator
$$\frac{1}{(1+x^2)^2}=\frac{1+x^2}{(1+x^2)^2}-\frac{x^2}{(1+x^2)}$$
The integral of the first term we know because it has a lower exponent and for the second we can write it as $$\frac{1}{2}x\frac{2x}{(1+x)^2}$$ and integrate by parts
$$\frac{1}{2}\int x\frac{2x}{(1+x^2)^2}=\frac{1}{2}x\frac{1}{(1+x^2)}+\frac{1}{2}\int\frac{1}{(1+x^2)}$$
which makes it boil down again to the integral with the smaller exponent, which we know.
Adding the two parts we get the answer $$\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}-\frac{1}{2}\arctan(x)=\frac{1}{2}\arctan(x)-\frac{1}{2}x\frac{1}{(1+x^2)}$$
A: Another way is to substitute $x=\tan(\theta)$. Then it boils down to 
$$\int \cos^2\theta d\theta = \theta/2 + \sin(2\theta)/2 + C$$
Now use $$\sin(2\theta) = 2\sin(\theta)\cos(\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)}$$
Now put $\theta=\tan^{-1}x$ and you are done.
A: The easy approach: Add them to your integral table...
$$ \int{\frac{a}{{(a^2+x^2)}^2}dx=}\ \left(\frac{1}{{2a}^2}\right)arctan(x/a)+\left(\frac{1}{2a}\right)\frac{x}{a^2+x^2}\ +C$$
set a = 1:
$$\int{\frac{1}{{(1+x^2)}^2}dx}=\frac{1}{2}\left(arctan(x)+\frac{x}{1+x^2}\right)+C$$
If you need to prove this yourself, you can obtain the result above  by taking the derivative of the following standard integral with respect a:
$$\int \frac{1}{a^{2}+x^{2}} dx = \frac{1}{a} arctan(x/a)$$
