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प्रश्न
After using \[t=\tan^{-1}x\] in \[\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx\], which integral results?
पर्याय
\[\int_{0}^{\frac{\pi}{4}}t\,dt\]
\[\int_{0}^{1}t\,dt\]
\[\int_{0}^{\frac{\pi}{4}}\frac{1}{1+t^2}\,dt\]
\[\int_{0}^{\frac{\pi}{4}}\tan t\,dt\]
MCQ
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उत्तर
The factor \[\frac{1}{1+x^2}\,dx\] becomes \[dt\], while \[\tan^{-1}x\] becomes \[t\]. The new limits are \[0\] and \[\frac{\pi}{4}\].
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