Question

If $\underset{ a \rightarrow \infty} {\text{Lim}} \frac{1}{ a } \int\limits_0^{\infty} \frac{ x ^2+ ax +1}{1+ x ^4} \cdot \tan ^{-1}\left(\frac{1}{ x }\right) dx$ 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=\int\limits_0^{\infty} \frac{x^2+a x+1}{1+x^4} \cdot \tan ^{-1}\left(\frac{1}{x}\right) d x$
put $x=1 / t$ and adding we get $$ (using $\tan ^{-1}(x)+\cot ^{-1} x=\pi / 2$ )
$I =\frac{\pi}{4} \int\limits_0^{\infty} \frac{\left( x ^2+1\right)+ ax }{1+ x ^4} dx =\frac{\pi}{4}\left[\int\limits_0^{\infty} \frac{\left( x ^2+1\right)}{1+ x ^4} dx + a \int\limits_0^{\infty} \frac{ xdx }{1+ x ^4}\right]=\frac{\pi}{4}\left[\frac{\pi}{2 \sqrt{2}}+\frac{ a \pi}{4}\right]=\left[\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi^2 a }{16}\right]$
$\therefore l=\underset { a \rightarrow \infty}{\text{Lim}} \frac{1}{ a }\left[\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi^2 a }{16}\right]=\underset { a \rightarrow \infty}{\text{Lim}} \left[\frac{\pi^2}{(8 \sqrt{2}) a }+\frac{\pi^2}{16}\right]=\frac{\pi^2}{16} \Rightarrow k =16$