# Evaluate $\int_0^\infty \frac{(1-x^2) \arctan x^2}{1+4x^2+x^4} \,{\rm d} x$

Evaluate $$\int_0^\infty \frac{(1-x^2) \arctan x^2}{1+4x^2+x^4} \,{\rm d} x$$ using any analytic method taught to you.

Edit: This is not a homework problem, I simply want to learn how to evaluate such a kind of integral. I tried simplifying the denominator and used a trig substitution, but I failed. I tried subbing x = (tan(a))^1/2 but ended up with a messy trig integral that I could not evaluate.

Wolfram Alpha's answer on my try with the trig sub

• Where is your attempt? – Sigma6RPU Aug 15 '16 at 0:06
• Read my comment. I tried using x = root(tan(a)), but got a really messy trig integral. – Why Do You Care Aug 15 '16 at 0:09
• Yes, I edited it so people like you wouldn't down vote my post. – Why Do You Care Aug 15 '16 at 0:12
• What was the messy trig integral? That substitution was the first to come to my mind as well. – MathematicsStudent1122 Aug 15 '16 at 0:13
• @mathreadler I tried that substitution before, but the integrand does not contain a cx term, where c is a constant. – Why Do You Care Aug 15 '16 at 0:43

## 1 Answer

According to wolfram alpha,

$$\int \frac{1-x^2}{1+4x^2+x^4} dx = \frac{-(\sqrt{3}-3)\sqrt{2+\sqrt{3}} \tan^{-1}(\frac{x}{\sqrt{2-\sqrt{3}}}) - \sqrt{2-\sqrt{3}}(3+\sqrt{3})\tan^{-1}(\frac{x}{\sqrt{2+\sqrt{3}}})}{2\sqrt{3}},$$

which can be obtained by doing a partial fraction decomposition.

It should be clear that,

$$\frac{d}{dx} \tan^{-1}(x^2) = \frac{2x}{1+x^4}$$

This means that the integral can be done by performing an integration by parts where $U=\tan^{-1}(x^2)$ and $dV=(1-x^2)/(1+4x^2+x^4)dx$.

The result is quite complicated and I will not type it all here; but it will have integrals of the form,

$$\int_0^\infty \tan^{-1}(\frac{x}{\sqrt{2-\sqrt{3}}})\frac{x}{x^4+1} dx =\pi^2/12,$$

$$\int_0^\infty \tan^{-1}(\frac{x}{\sqrt{2+\sqrt{3}}})\frac{x}{x^4+1} dx =\pi^2/24,$$

these final values I got from wolfram alpha/mathematica.

You should be able to compute the integral with the information I have provided here.

• @WhyDoYouCare, no problem it was my pleasure. – Spencer Aug 15 '16 at 4:49