Question

If $\underset{a\to \infty }{\mathrm{Lim}}\frac{1}{a}\underset{0}{\overset{\infty }{\int }}\frac{x^2+ax+1}{1+x^4}\cdot {\tan }^{-1}\left(\frac{1}{x}\right)dx$ is equal to $\frac{{\pi }^2}{k}$ where $k\in N$ equals

• A. 4
• B. 8
• C. 16
• D. 32

Answer 1

Answer: (C)

Let $I=\underset{0}{\overset{\infty }{\int }}\frac{x^2+ax+1}{1+x^4}\cdot {\tan }^{-1}\left(\frac{1}{x}\right)dx$
put $x=1/t$ and adding we get (using ${\tan }^{-1}(x)+{\cot }^{-1}x=\pi /2$ )
$I=\frac{\pi }{4}\underset{0}{\overset{\infty }{\int }}\frac{\left(x^2+1\right)+ax}{1+x^4}dx=\frac{\pi }{4}\left[\underset{0}{\overset{\infty }{\int }}\frac{\left(x^2+1\right)}{1+x^4}dx+a\underset{0}{\overset{\infty }{\int }}\frac{xdx}{1+x^4}\right]=\frac{\pi }{4}\left[\frac{\pi }{2\sqrt{2}}+\frac{a\pi }{4}\right]=\left[\frac{{\pi }^2}{8\sqrt{2}}+\frac{{\pi }^{2}a}{16}\right]$
$\therefore l=\underset{a\to \infty }{\mathrm{Lim}}\frac{1}{a}\left[\frac{{\pi }^2}{8\sqrt{2}}+\frac{{\pi }^{2}a}{16}\right]=\underset{a\to \infty }{\mathrm{Lim}}\left[\frac{{\pi }^2}{(8\sqrt{2})a}+\frac{{\pi }^2}{16}\right]=\frac{{\pi }^2}{16}\Rightarrow k=16$