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प्रश्न

\[\int e^\sqrt{x} \text{ dx }\]
योग
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उत्तर

\[\text{ Let I } = \int e^\sqrt{x}  \text{ dx }\]
\[ = \int\sqrt{x} \cdot \frac{e^\sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2 dt\]
\[ \therefore I = 2\int t_{} \cdot {e^t}_{} dt\]
`  " Taking t as the first function and e"^t" as the second function " .  `
\[ = 2\left[ t\int e^t dt - \int\left\{ \frac{d}{dt}\left( t \right)\int e^t dt \right\}dt \right] \]
\[ = 2\left[ t \cdot e^t - \int1 \cdot e^t dt \right] + C . . . (1)\]
\[\text{Substituting the value of t in eq}  \text{ (1) }\]
\[ = 2\left[ \sqrt{x} \text{ e }^\sqrt{x} - e^\sqrt{x} \right] + C\]
\[ = 2 \text{ e}^\sqrt{x} \left( \sqrt{x} - 1 \right) + C\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.25 | Q 22 | पृष्ठ १३३

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