$$\text{If }\int_{0}^{\infty}\frac{dx}{1+x^2+x^4}=\frac{\pi \sqrt{n}}{2n},\text{then } n=$$
$$$$
$$\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}$$
Using partial fraction decomposition, I ended with $n=3$, that is option $\text{C}$. However, it took me about $8$ minutes, which is not feasible for an MCQ test, where the average time to solve a problem is $3$ minutes, and no calculator is allowed.
I believe that there is something obvious for some of you. If there is nothing obvious, then there must be a simpler and a quicker way than partial fraction decomposition.
This post is voted to be closed and considered as [duplicate] but it is not. To prove is different than evaluating $n$ with possibly a cliver and a fast way. To prove, I can use partial fraction decomposition. Some users DO NOT really read the post properly!
Please share your thoughts. THANKS!
Best Answer
Note that $I=\int_{0}^{\infty}\frac{1}{1+x^2+x^4}dx\overset{x\to \frac1x}= \int_{0}^{\infty}\frac{x^2}{1+x^2+x^4}dx $. Then $$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} $$