Advertisements
Advertisements
प्रश्न
\[\int x e^x \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\int x e^x \text{ dx }\]
` "Taking x as the first function and e"^x " as the second function"`
\[ = x\int e^x dx - \int\left\{ \frac{d}{dx}\left( x \right)\int e^x dx \right\}dx\]
\[ = x e^x - \int1\left( e^x \right)dx\]
\[ = x e^x - e^x + C\]
\[ = \left( x - 1 \right) e^x + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{x^6 + 1}{x^2 + 1} dx\]
\[\int\frac{\text{sin} \left( x - a \right)}{\text{sin}\left( x - b \right)} dx\]
\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]
` ∫ {sec x "cosec " x}/{log ( tan x) }` dx
\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int\sqrt{1 + e^x} . e^x dx\]
\[\int \sin^5\text{ x }\text{cos x dx}\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int\frac{1}{\sqrt{x} + \sqrt[4]{x}}dx\]
\[\int \sec^4 2x \text{ dx }\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{x^2 + x + 1}{x^2 - x} dx\]
\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{ dx }\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{ dx }\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
Find \[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to
\[\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx =\]
\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\]
\[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int\frac{\log x}{x^3} \text{ dx }\]
\[\int\frac{x^2}{\sqrt{1 - x}} \text{ dx }\]
