मराठी

3 ∫ 0 3 X + 1 X 2 + 9 D X = - Mathematics

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प्रश्न

\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

पर्याय

  • \[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{2} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{6} + \log\left( 2\sqrt{2} \right)\]
  • \[\frac{\pi}{3} + \log\left( 2\sqrt{2} \right)\]

MCQ
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उत्तर

\[\frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]

\[\text{We have}, \]
\[I = \int_0^3 \frac{3x + 1}{x^2 + 9} d x\]
\[I = \int_0^3 \frac{3x}{x^2 + 9}dx + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ I_1 = \int_0^3 \frac{3x}{x^2 + 9}dx and I_2 = \int_0^3 \frac{1}{x^2 + 9}dx\]
\[\text{Putting} x^2 + 9 = t in I_1 \]
\[ \Rightarrow 2x\ dx = dt\]
\[ \Rightarrow x\ dx = \frac{dt}{2}\]
\[When\ x \to 0; t \to 9\]
\[and\ x \to 3; t \to 18\]
\[ \therefore I = \int_9^{18} \frac{3 dt}{2 t} + \int_0^3 \frac{1}{x^2 + 9}dx\]
\[ = \frac{3}{2} \int_9^{18} \frac{dt}{t} + \int_0^3 \frac{1}{x^2 + 3^2}dx\]
\[ = \frac{3}{2} \left[ \log\left( t \right) \right]_9^{18} + \frac{1}{3} \left[ \tan^{- 1} \left( \frac{x}{3} \right) \right]_0^3 \]
\[ = \frac{3}{2}\left[ \log18 - \log9 \right] + \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{3}{2}\left[ \log\frac{18}{9} \right] + \frac{\pi}{12}\]
\[ = \frac{3}{2}\left[ \log 2 \right] + \frac{\pi}{12}\]
\[ = \log\left( \sqrt{8} \right) + \frac{\pi}{12}\]
\[ = \log\left( 2\sqrt{2} \right) + \frac{\pi}{12}\]
\[ = \frac{\pi}{12} + \log\left( 2\sqrt{2} \right)\]

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Definite Integrals
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पाठ 20: Definite Integrals - MCQ [पृष्ठ ११८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 16 | पृष्ठ ११८

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