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Grade: 12th pass
        
q. 12 plese solve it
one year ago

Answers : (4)

Harsh
29 Points
							
Please post questions properly. Attachment cant be seen properly. Please post right image which is readable.
one year ago
Piyush Kumar Behera
385 Points
							
yes  the attachment couldnot be seen properly,please again repost the question ,so that it can be read and can be solved.
one year ago
jagdish singh singh
173 Points
							
\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 $I = \lim_{a\rightarrow \infty}\int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx =\frac{\pi^2}{16}$
one year ago
jagdish singh singh
173 Points
							
\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 $I = \lim_{a\rightarrow \infty}\int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx =\frac{\pi^2}{16}$
…......................................................................................................
one year ago
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