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प्रश्न
विकल्प
− xex + C
xex + C
− xe−x + C
xe−x + C
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उत्तर
− xe−x + C
\[\int \left( x - 1 \right)_I {e^{- x}}_{II} dx\]
\[ = \left( x - 1 \right)\int e^{- x} dx - \int\left\{ \frac{d}{dx}\left( x - 1 \right)\int e^{- x} dx \right\}dx\]
\[ = \left( x - 1 \right) \cdot e^{- x} \left( - 1 \right) - \int1 \cdot e^{- x} \times - 1 dx\]
\[ = - \left( x - 1 \right) e^{- x} + \int e^{- x} dx\]
\[ = \left( 1 - x \right) e^{- x} + \frac{e^{- x}}{- 1} + C\]
\[ \Rightarrow \left( 1 - x - 1 \right) e^{- x} + C\]
\[ = - x e^{- x} + C\]
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` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
Evaluate the following integrals:
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
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