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प्रश्न
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उत्तर
\[Let\ I = \int_0^1 \frac{\tan^{- 1} x}{1 + x^2} d\ x . Then, \]
\[Let\ \tan^{- 1} x = t . Then, \frac{1}{1 + x^2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = 1, t = \frac{\pi}{4}\]
\[ \therefore I = \int_0^\frac{\pi}{4} t dt\]
\[ \Rightarrow I = \left[ \frac{t^2}{2} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \frac{\pi^2}{32}\]
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