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.

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Partial fractions doesn't work when the two factors are the same; the expression is already simplified. Have you tried trig substitution? – Qiaochu Yuan Apr 29 '11 at 21:19
Forget about where your book says you should use it. Understand why trig substitution works in the situations where it does, and what identities it takes advantage of, and you'll see how to use it here. – MartianInvader Apr 29 '11 at 21:29
Whenever I see $1+x^2$ in an integral I think of substituting $x=\tan \theta$. Doesn't always work... – Ross Millikan Apr 29 '11 at 21:31
@Virtuoso: Of course you know how to integrate $1/(1+x^2)$. But temporarily pretend you don't, and try to integrate it by parts, using $du=dx$ and $v=1/(1+x^2)$. Something nice will happen. (But probably you will find trig substitution more natural!) – André Nicolas Apr 29 '11 at 21:32
@Ross: ...but whenever it does, the problem vastly simplifies. :) Yet another way of going about it if one insists on doing a partial fraction decomposition would be to consider $x^2+1=(x+i)(x-i)$... – J. M. Apr 30 '11 at 0:07

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.

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do you have any sort of thoery/reference on why you are allowed to do this? i didn't learn how to do it this way. – chaver Apr 29 '11 at 21:40
It's simply a $u$-substitution. You say you have already learned trig substitution, and this is just an application of it. The wikipedia page might be helpful, but this is one of the standard methods taught in any Calc 1 class. – yunone Apr 29 '11 at 21:43
I'd say it's been in Calc 2 in all of my experiences. – Graphth Apr 30 '11 at 1:37
@Virtuoso: Well written notes on trig substitution can be found at math.stackexchange.com/questions/60444/… – JavaMan Aug 31 '11 at 19:05

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 $\dfrac{1}{1+x^{2}}.$

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Nice algorithm for solving this integral. – Pedro Tamaroff Apr 28 '12 at 18:48
@PeterTamaroff Thanks! – Américo Tavares Apr 28 '12 at 20:09

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.

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Pretty! – André Nicolas Apr 30 '11 at 0:28
@user6312: thank you... – Fabian Apr 30 '11 at 10:59

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.

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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...

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Put $x=\tan \theta$. This will give you $$\int \cos^2\theta\ d\theta$$ Hope you can solve it now.

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Thank you very much , I would've never thought about that substition ! – Victor May 14 '14 at 12:09

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.

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$$\frac{1}{(1+x^2)^2} = \left(\frac{1}{1+x^2}-\frac{x^2}{(1+x^2)^2}\right)= \left(\frac{1}{1+x^2}+\frac x2\left(\frac{1}{ 1+x^2 }\right)'\right)$$$$=\left(\frac{1}{1+x^2}+\left(\frac 12 \frac {x}{1+x^2}\right)'-\frac 12 \frac {1}{1+x^2}\right) = \frac 12 \left(\arctan x+\frac{x}{1+x^2}\right)'.$$

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