# q. 12 plese solve it

Harsh
29 Points
6 years ago
Piyush Kumar Behera
436 Points
6 years ago
yes  the attachment couldnot be seen properly,please again repost the question ,so that it can be read and can be solved.
jagdish singh singh
173 Points
6 years ago
$\hspace{-0.70 cm}Let I = \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx\cdots \cdots (1)\\\\\\ put x=\frac{1}{t}\;, Then dx = -\frac{1}{t^2}dt and changing limits, We get \\\\\\ I = \int^{\infty}_{0}\frac{t^2+at+1}{1+t^4}\cdot \tan^{-1}(t)dt= \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}(x)dx\cdots (2)\\\\\\$

$\hspace{-0.70 cm}2I = \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\left(\cot^{-1}(x)+\tan^{-1}(x)\right)dx \\\\\\ 2I=\frac{\pi}{2} \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}dx\;, bcz \cot^{-1}(x)+\tan^{-1}(x)=\frac{\pi}{2},x>0\\\\\\ So I = \frac{\pi}{2}\underbrace{\int^{\infty}_{0}\frac{x^2+1}{x^4+1}dx}_{\bf{finite}}+\frac{a\cdot \pi}{8}\int^{\infty}_{0}\frac{(x^2)'}{1+(x^2)^2}dx = \frac{a\cdot \pi^2}{16}$

$\hspace{-0.70 cm}So I = \lim_{a\rightarrow \infty}\frac{1}{a}\int^{\infty}_{0}\frac{1+x^2}{1+x^4}dx+\lim_{a\rightarrow \infty}\frac{a\cdot \pi^2}{16 \cdot a}\\\\\\ So we get I = \lim_{a\rightarrow \infty}\int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx =\frac{\pi^2}{16}$
jagdish singh singh
173 Points
6 years ago
$\hspace{-0.70 cm}Let I = \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx\cdots \cdots (1)\\\\\\ put x=\frac{1}{t}\;, Then dx = -\frac{1}{t^2}dt and changing limits, We get \\\\\\ I = \int^{\infty}_{0}\frac{t^2+at+1}{1+t^4}\cdot \tan^{-1}(t)dt= \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}(x)dx\cdots (2)\\\\\\$

$\hspace{-0.70 cm}2I = \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\left(\cot^{-1}(x)+\tan^{-1}(x)\right)dx \\\\\\ 2I=\frac{\pi}{2} \int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}dx\;, bcz \cot^{-1}(x)+\tan^{-1}(x)=\frac{\pi}{2},x>0\\\\\\ So I = \frac{\pi}{2}\underbrace{\int^{\infty}_{0}\frac{x^2+1}{x^4+1}dx}_{\bf{finite}}+\frac{a\cdot \pi}{8}\int^{\infty}_{0}\frac{(x^2)'}{1+(x^2)^2}dx = \frac{a\cdot \pi^2}{16}$

$\hspace{-0.70 cm}So I = \lim_{a\rightarrow \infty}\frac{1}{a}\int^{\infty}_{0}\frac{1+x^2}{1+x^4}dx+\lim_{a\rightarrow \infty}\frac{a\cdot \pi^2}{16 \cdot a}\\\\\\ So we get I = \lim_{a\rightarrow \infty}\int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx =\frac{\pi^2}{16}$
…......................................................................................................