Advertisements
Advertisements
Question
\[\int x e^x \text{ dx }\]
Sum
Advertisements
Solution
\[\int x e^x \text{ dx }\]
` "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
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int\frac{x^2 + x + 5}{3x + 2} dx\]
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]
` ∫ sin 4x cos 7x dx `
\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int\frac{e^{m \tan^{- 1} x}}{1 + x^2} dx\]
\[\int\sqrt {e^x- 1} \text{dx}\]
\[\int\frac{x^2 + 3x + 1}{\left( x + 1 \right)^2} dx\]
\[\int \cot^6 x \text{ dx }\]
\[\int \sin^3 x \cos^5 x \text{ dx }\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int\frac{1}{\sqrt{3} \sin x + \cos x} dx\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int2 x^3 e^{x^2} dx\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
\[\int x \sin^3 x\ dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]
\[\int\frac{\cos x}{\left( 1 - \sin x \right)^3 \left( 2 + \sin x \right)} dx\]
\[\int\frac{2x + 1}{\left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{4 x^4 + 3}{\left( x^2 + 2 \right) \left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
\[\int \sec^4 x\ dx\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int\sqrt{x^2 - a^2} \text{ dx}\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .
\[\int \left( e^x + 1 \right)^2 e^x dx\]
