If $\int_{0}^{\infty}\frac{dx}{1+x^2+x^4}=\frac{\pi \sqrt{n}}{2n}$, then $n=$

Question

If $\int_{0}^{\infty}\frac{dx}{1+x^2+x^4}=\frac{\pi \sqrt{n}}{2n}$, then $n=$

$\text{A) }1 \space \space \space \space \space\text{B) }2 \space \space \space \space \space\text{C) }3 \space \space \space \space \space\text{D) }4 \space \space \space \space \space\text{E) }\text{None of the given options}$

Using partial fraction decomposition, I ended with $n=3$, that is option $\text{C}$. However, it took me about $8$ minutes, which is not feasible for an MCQ test, where the average time to solve a problem is $3$ minutes, and no calculator is allowed.

I believe that there is something obvious for some of you. If there is nothing obvious, then there must be a simpler and a quicker way than partial fraction decomposition.

This post is voted to be closed and considered as [duplicate] but it is not. To prove is different than evaluating $n$ with possibly a cliver and a fast way. To prove, I can use partial fraction decomposition. Some users DO NOT really read the post properly!

Please share your thoughts. THANKS!

Answer 1

Note that $I=\int_{0}^{\infty}\frac{1}{1+x^2+x^4}dx\overset{x\to \frac1x}= \int_{0}^{\infty}\frac{x^2}{1+x^2+x^4}dx $. Then $$I= \frac12\int_{0}^{\infty}\frac{1+x^2}{1+x^2+x^4}dx= \frac12\int_{0}^{\infty}\frac{d(x-\frac1{x})}{(x-\frac1{x})^2+3}dx\overset{t=x-\frac1x}=\frac\pi{2\sqrt3} $$

Answer 2

There's a very quick way to do it with complex analysis:
Consider the semicircular contour integral $$\int_{\gamma} \frac{dz}{1 + z^2 + z^4}$$ around the poles $\omega = e^{i\pi / 3}$ and $\omega^2 = e^{2i\pi / 3}$. As the radius of the semicircle increases, the integral along the curved part decreases in magnitude, and the integral approaches $$\int_{-\infty}^\infty \frac{dx}{1 + x^2 + x^4},$$ twice our desired integral. By Cauchy's residue theorem, this will be $2\pi i$ times the sum of the residues at $\omega$ and $\omega^2$. By L'Hospital's rule, the residue at $\omega$ is $$\lim_{z \to \omega} \frac{z - \omega}{1 + z^2 + z^4}dz = \lim_{z \to \omega}\frac{1}{2z + 4z^3} = \frac{1}{\sqrt3i - 3} = -\frac{\sqrt3 i + 3}{12}.$$ Similarly, the residue at $\omega^2$ is $$-\frac{\sqrt3 i - 3}{12}.$$ Thus, $$\int_{-\infty}^\infty \frac{dx}{1 + x^2 + x^4} = 2\pi i\left(-\frac{\sqrt3 i + 3}{12} + -\frac{\sqrt3 i - 3}{12}\right) = \frac{\pi\sqrt3}{3}.$$ This means the desired integral equals $\pi \sqrt3 / 6$, so the answer is $\boxed 3$.

Answer 3

$$ \begin{aligned} \int_{0}^{\infty} \frac{1}{1+x^{2}+x^{4}} d x &=\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}+1} d x \\ &=\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}+1} d x \\ &=\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+3}-\int_{0}^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-1}\right] \\ &=\frac{1}{2}\left[\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{1}{2} \ln \left|\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right|\right]_{0}^{\infty} \\ &=\frac{1}{2}\left[\frac{1}{\sqrt{3}}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)\right] \\ &=\frac{\pi \sqrt{3}}{6} \end{aligned} $$

Hence the answer is

$$\boxed{C) :n=3.}$$