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1 ∫ 0 1 2 X 2 + X + 1 D X

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प्रश्न

\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]
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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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अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 40 | पृष्ठ १७

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