Integral of $\frac{1}{(1+x^2)^2}$ I am in the middle of a problem and having trouble integrating the following integral:
$$\int_{-1}^1\frac1{(1+x^2)^2}\mathrm dx$$
I tried doing partial fractions and got:
$$1=A(1+x^2)+B(1+x^2)$$
I have no clue how to solve this since it is obvious there is no way to cancel out either $A$ or $B$ to get the other variable. Please guide me.
Thank you.
 A: Hint:
On this page, you find the reduction formula
$$\int \frac{dx}{(x^2 +m^2)^k} = x\frac1{2 m^2 (k-1) (x^2 +m^2)^{k-1}}+
\frac{2k-3}{2m^2 (k-1)} \int \frac{dx}{(x^2 + m^2)^{k-1}}$$
which you can use for your integral...
A: Try making a substitution $x=\tan u$. Notice then that
$$
(1+x^2)^2=(1+\tan^2 u)^2=(\sec^2 u)^2=\sec^4 u
$$
and
$$
dx=\sec^2 u\ du
$$
So the indefinite integral is now
$$
\int\frac{1}{\sec^2 u}du=\int\cos^2 u\ du.
$$
This new integrand should be easier to integrate. Just remember to change your limits to get the proper evaluation.
A: The integrand is already a partial fraction, as pointed out by Qiaochu Yuan.

If we add and subtract $x^2$ in the numerator, we can integrate the first integral immediately 
$$\begin{eqnarray*}
\int \frac{1}{\left( 1+x^{2}\right) ^{2}}dx &=&\int \frac{1}{1+x^{2}}dx-\int \frac{x^{2}}{\left( 1+x^{2}\right) ^{2}}dx
\\
&=&\arctan x-\int x\frac{x}{\left( 1+x^{2}\right) ^{2}}dx
\end{eqnarray*}$$
and the second integral by parts:
$$\begin{eqnarray*}
\int x\frac{x}{\left( 1+x^{2}\right) ^{2}}dx &=&x\left( -\frac{1}{2\left(
1+x^{2}\right) }\right) +\int \frac{1}{2\left( 1+x^{2}\right) }dx \\
&=&-\frac{x}{2\left( 1+x^{2}\right) }+\frac{1}{2}\arctan x.
\end{eqnarray*}$$

Added: by applying this method $n-1$ times, we can reduce the integration of the function $f(x)=\dfrac{1}{\left( 1+x^{2}\right) ^{n}}$ to the integration of $\dfrac{1}{1+x^{2}}.$
A: $$\frac{1}{(1+x^2)^2} =  \left(\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}\right)= \left(\frac{1}{1+x^2}+\frac x2\left(\frac{1}{ 1+x^2 }\right)'\right)$$$$=\left(\frac{1}{1+x^2}+\left(\frac 12 \frac {x}{1+x^2}\right)'-\frac 12 \frac {1}{1+x^2}\right) = \frac 12 \left(\arctan x+\frac{x}{1+x^2}\right)'.$$
A: Put $x=\tan \theta$. This will give you $$\int \cos^2\theta\ d\theta $$ Hope you can solve it now.
A: If want to solve the integral using partial integration (as indicated in the question), you can break the degeneracy of the root of the polynomial in the denominator which hinders you from applying partial fraction expansion. I.e., you write the more general integral
$$\int_{-1}^1 \frac{dx}{(1+x^2)(a^2+x^2)}$$ and obtain the result which you want by sending $a\to 1$ in the end. As you want to integral $x$ from -1 to 1 you should keep $a>1$ and send it to 1 from above.
Using partial fraction expansion
$$ \frac{1}{(1+x^2)(a^2+x^2)} = \frac1{(a^2-1)}\left[\frac1{(1+x^2)} - \frac1{(a^2 +x^2)}\right]$$ you can reduce it onto more elementary integrals which you can compute easily. The result reads
$$\int_{-1}^1 \frac{dx}{(1+x^2)(a^2+x^2)} = \frac{\arctan(x)-\arctan(x/a)/a }{a^2-1}\Bigr|_{x=-1}^1 \,.$$
Sending $a\to 1$, you obtain (with de l'Hôpital) the result.
A: Another approach to this one is differentiation under the integral sign.
Hint:
If you let $$I(a) = \int_{-1}^1\frac{1}{a+x^2}\mathrm d x,$$ then the value that you want to compute is $$\left. -\frac{\partial}{\partial a} I(a)\right|_{a=1}.$$
Now, $I(a)$ can be calculated using the substitution $x=\sqrt{a} \tan \theta$, as mentioned in the comments.
EDIT: Note that like in Américo Tavares' answer, this readily generalises to higher powers of the integrand, just by differentiating repeatedly, and being careful about signs.
A: You can use by parts to express $\int\frac{dx}{1+x^2}$, whose closed form you know, in terms of your unknown integral and treat it like an equation. Set
$u'=1$,$u=x$,$v=\frac{1}{1+x^2}$,$v'=-\frac{2x}{(1+x^2)^2}$
and you get
$\int\frac{dx}{1+x^2}=\frac{x}{1+x^2}+2\ln{|1+x^2|}-2\int\frac{1}{(1+x^2)^2}=\arctan{x}$
and then solve for your integral.
A: $$
\begin{aligned}
\int \frac{1}{\left(1+x^{2}\right)^{2}} d x\stackrel{IBP}{=} &-\frac{1}{2} \int \frac{1}{x} d\left(\frac{1}{1+x^{2}}\right) \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \int\left(\frac{1}{1+x^{2}}-\frac{1}{x^{2}}\right) d x \\=&-\frac{1}{2 x\left(1+x^{2}\right)}+\frac{1}{2} \tan ^{-1} x+\frac{1}{2 x}+C\\
=&\frac{1}{2}\left(\frac{x}{1+x^{2}}+\tan ^{-1} x\right)+C
\end{aligned}
$$
A: $$\dfrac{d(x(x^2+a^2)^m)}{dx}=(x^2+a^2)^m+2m(x^2+a^2)^{m-1}(x^2+a^2-a^2)$$
Integrate both sides with respect to $x,$
$$x(x^2+a^2)^m=I(m)(1+2m)-2ma^2I(m-1)$$
where $$I(n)=\int(x^2+a^2)^n\ dx$$
Set $m=-1$
