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प्रश्न
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उत्तर
\[Let\ I = \int_0^1 \frac{1}{2 x^2 + x + 1} d\ x . Then, \]
\[I = \frac{1}{2} \int_0^1 \frac{1}{x^2 + \frac{x}{2} + \frac{1}{2}} d x\]
\[I = \frac{1}{2} \int_0^1 \frac{1}{\left( x^2 + \frac{x}{2} + \frac{1}{16} \right) - \frac{1}{16} + \frac{1}{2}} d\ x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^1 \frac{1}{\left( x + \frac{1}{4} \right)^2 + \frac{7}{16}} dx\]
\[ \Rightarrow I = \frac{1}{2} \times \frac{4}{\sqrt{7}} \left[ \tan^{- 1} \left( \frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}} \right) \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{\sqrt{7}}\left( \tan^{- 1} \frac{5}{\sqrt{7}} - \tan^{- 1} \frac{1}{\sqrt{7}} \right)\]
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