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