English

I N T X E X 2 D X - Mathematics

Advertisements
Advertisements

Question

\[\  ∫    x   \text{ e}^{x^2} dx\]
Sum
Advertisements

Solution

\[\int x . e^{x^2} dx\]
\[\text{Let x}^2 = t\]
\[ \Rightarrow \text{2x dx} = dt\]
\[ \Rightarrow \text{x dx} = \frac{dt}{2}\]
\[Now, \int x . e^{x^2} dx\]
\[ = \frac{1}{2}\int e^t dt\]
\[ = \frac{1}{2} e^t + C\]
\[ = \frac{1}{2} e^{x^2} + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 59]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 59 | Page 59

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]

\[\int\frac{\sin^2 x}{1 + \cos x}   \text{dx} \]

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


` ∫  1/ {1+ cos   3x}  ` dx


` ∫   sin x  \sqrt (1-cos 2x)    dx `

 


\[\int\frac{1}{\sqrt{1 - \cos 2x}} dx\]

\[\int\sqrt {e^x- 1}  \text{dx}\] 

\[\int\frac{1}{a^2 x^2 + b^2} dx\]

\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]

 


\[\int\frac{1}{2 x^2 - x - 1} dx\]

\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]

\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]

\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]

\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]

\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]

\[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} dx\]

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6}\text{  dx }\]

\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]

\[\int\frac{x}{\sqrt{x^2 + x + 1}} \text{ dx }\]

\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]

\[\int\frac{x + \sin x}{1 + \cos x} \text{ dx }\]

\[\int {cosec}^3 x\ dx\]

\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]

\[\int e^x \left( \cos x - \sin x \right) dx\]

\[\int e^x \left( \frac{\sin x \cos x - 1}{\sin^2 x} \right) dx\]

\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{  dx }\]

\[\int x\sqrt{x^2 + x} \text{  dx }\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]

\[\int\frac{x^2 - 3x + 1}{x^4 + x^2 + 1} \text{ dx }\]

\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]

\[\int\frac{1}{1 + \tan x} dx =\]

\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

\[\int \sec^2 x \cos^2 2x \text{ dx }\]

\[\int \text{cosec}^2 x \text{ cos}^2 \text{  2x  dx} \]

\[\int \tan^5 x\ dx\]

\[\int\frac{1}{2 + \cos x} \text{ dx }\]


\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×