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प्रश्न

\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
बेरीज
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उत्तर

\[\text{ Let I }= \int e^x \left( \frac{x - 1}{2 x^2} \right)dx\]

\[ = \frac{1}{2}\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)dx\]

\[\text{ here }\frac{1}{x} = f(x) \text{ Put  e}^x f(x) = t\]

\[ \Rightarrow - \frac{1}{x^2} = f'(x)\]

\[\text{ let e}^x \frac{1}{x} = t\]

\[\text{ Diff both sides w . r . t x}\]

\[\left( e^x \frac{1}{x} + e^x \frac{- 1}{x^2} \right) = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left( \frac{1}{x} - \frac{1}{x^2} \right)dx = dt\]

\[ \therefore I = \frac{1}{2}\int dt\]

\[ = \frac{t}{2} + C\]

\[ = \frac{e^x}{2x} + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.26 | Q 5 | पृष्ठ १४३

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