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ADNAN MUHAMMED

Grade 12,

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q. 12 plese solve it

q. 12 plese solve it

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Grade:12th pass

4 Answers

Harsh
29 Points
7 years ago
Please post questions properly. Attachment cant be seen properly. Please post right image which is readable.
Piyush Kumar Behera
417 Points
7 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
7 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
7 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 $I = \lim_{a\rightarrow \infty}\int^{\infty}_{0}\frac{x^2+ax+1}{1+x^4}\cdot \tan^{-1}\left(\frac{1}{x}\right)dx =\frac{\pi^2}{16}$
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