Advertisements
Advertisements
प्रश्न
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
पर्याय
− xex + C
xex + C
− xe−x + C
xe−x + C
MCQ
उत्तर
− xe−x + C
\[\int \left( x - 1 \right)_I {e^{- x}}_{II} dx\]
\[ = \left( x - 1 \right)\int e^{- x} dx - \int\left\{ \frac{d}{dx}\left( x - 1 \right)\int e^{- x} dx \right\}dx\]
\[ = \left( x - 1 \right) \cdot e^{- x} \left( - 1 \right) - \int1 \cdot e^{- x} \times - 1 dx\]
\[ = - \left( x - 1 \right) e^{- x} + \int e^{- x} dx\]
\[ = \left( 1 - x \right) e^{- x} + \frac{e^{- x}}{- 1} + C\]
\[ \Rightarrow \left( 1 - x - 1 \right) e^{- x} + C\]
\[ = - x e^{- x} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\left( \sec^2 x + {cosec}^2 x \right) dx\]
\[\int\frac{\cos x}{1 + \cos x} dx\]
\[\int\frac{x + 3}{\left( x + 1 \right)^4} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
` ∫ sin 4x cos 7x dx `
Integrate the following integrals:
\[\int\text{sin 2x sin 4x sin 6x dx} \]
\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\]
\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{\cos^3 x}{\sqrt{\sin x}} dx\]
\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{\sec^2 x}{\sqrt{4 + \tan^2 x}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x + 1} \text{ dx }\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int x \text{ sin 2x dx }\]
\[\int x \cos^2 x\ dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int\frac{1}{x\left( x - 2 \right) \left( x - 4 \right)} dx\]
\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to
` \int \text{ x} \text{ sec x}^2 \text{ dx is equal to }`
\[\int\frac{1}{e^x + e^{- x}} dx\]
\[\int \tan^4 x\ dx\]
\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
\[\int \left( \sin^{- 1} x \right)^3 dx\]
\[\int\frac{x}{x^3 - 1} \text{ dx}\]