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2 ∫ 1 ( X − 1 X 2 ) E X D X

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प्रश्न

\[\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx\]
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उत्तर

\[Let\ I = \int_1^2 \left( \frac{x - 1}{x^2} \right) e^x\ d\ x . Then, \]
\[I = \int_1^2 \left( \frac{e^x}{x} - \frac{e^x}{x^2} \right) dx\]
\[ \Rightarrow I = \int_1^2 \frac{e^x}{x} dx - \int_1^2 \frac{e^x}{x^2} dx\]
\[\text{Integrating first term by parts}\]
\[I = \left\{ \left[ \frac{e^x}{x} \right]_1^2 - \int_1^2 \frac{- 1}{x^2} e^x dx \right\} - \int_1^2 \frac{e^x}{x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^x}{x} \right]_1^2 + \int_1^2 \frac{e^x}{x^2} dx - \int_1^2 \frac{e^x}{x^2} dx\]
\[ \Rightarrow I = \left[ \frac{e^x}{x} \right]_1^2 \]
\[ \Rightarrow I = \frac{e^2}{2} - e\]

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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 47 | पृष्ठ १७

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