English

∫ E X { F ( X ) + F ′ ( X ) } D X = - Mathematics

Advertisements
Advertisements

Question

\[\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx =\]
 

Options

  • ex f (x) + C

  • ex + (x)

  •  2ex f (x)

  •  ex − f (x)

MCQ
Advertisements

Solution

ex f (x) + C

 

\[\text{Let }I = \int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\}dx\]

\[\text{Putting }e^x f\left( x \right) = t\]

\[ \Rightarrow \left[ e^x \cdot f\left( x \right) + e^x f'\left( x \right) \right]dx = dt\]

\[ \therefore I = \int dt\]

\[ = t + C\]

\[ = e^x f\left( x \right) + C .............\left[ \because t = e^x f\left( x \right) \right]\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - MCQ [Page 202]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
MCQ | Q 27 | Page 202

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]

\[\int\frac{2 - 3x}{\sqrt{1 + 3x}} dx\]

\[\int \sin^2\text{ b x dx}\]

`  ∫  sin 4x cos  7x  dx  `

` ∫    cos  mx  cos  nx  dx `

 


\[\int\text{sin mx }\text{cos nx dx m }\neq n\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

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

\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]

\[\ \int\ x \left( 1 - x \right)^{23} dx\]

 


` ∫  tan^5 x   sec ^4 x   dx `

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

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

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

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

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

\[\int\frac{1}{4 \sin^2 x + 5 \cos^2 x} \text{ dx }\]

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

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

\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{  dx }\]

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

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

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

\[\int\sqrt{3 - x^2} \text{ dx}\]

\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]

\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]

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

Evaluate the following integral:

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

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

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

\[\int\text{ cos x  cos  2x   cos  3x  dx}\]


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

\[\int\sqrt{\text{ cosec  x} - 1} \text{ dx }\]

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


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

\[\int\frac{\log \left( \log x \right)}{x} \text{ dx}\]

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

Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×