English

∫ a B + C E X D X - Mathematics

Advertisements
Advertisements

Question

\[\int\frac{a}{b + c e^x} dx\]
Sum
Advertisements

Solution

\[\text{Let I} = \int\frac{a}{b + c e^x}dx\]
 ` "Dividing numerator and denominator by"   e^x `
\[ \Rightarrow I = \int\frac{a e^{- x}}{b e^{- x} + c}dx\]
\[Putting\ e^{- x} = t\]
\[ \Rightarrow - e^{- x} = \frac{dt}{dx}\]
\[ \Rightarrow e^{- x} dx = - dt\]
\[ \therefore I = \int\frac{- a}{bt + c}dt\]
\[ = \frac{- a}{b} \text{ln }\left| bt + c \right| + C \left[ \because \int\frac{1}{ax + b}dx = \frac{1}{a}\text{ln }\left| ax + b \right| + C \right]\]
\[ = \frac{- a}{b} \text{ln} \left| b e^{- x} + c \right| + C \left[ \because t = e^{- x} + C \right]\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.08 [Page 47]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.08 | Q 22 | Page 47

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

\[\int \left( \tan x + \cot x \right)^2 dx\]

\[\int\frac{1}{1 - \sin x} dx\]

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


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

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

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

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

\[\int \sin^4 x \cos^3 x \text{ dx }\]

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

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

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

\[\int\frac{x}{\sqrt{x^2 + 6x + 10}} \text{ dx }\]

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

\[\int\frac{1}{1 - \cot x} dx\]

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


\[\int e^\sqrt{x} \text{ dx }\]

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

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

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

\[\int\left( 2x + 3 \right) \sqrt{x^2 + 4x + 3} \text{  dx }\]

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

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

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

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

\[\int e^x \left( 1 - \cot x + \cot^2 x \right) dx =\]

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

The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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


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

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

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

\[\int\frac{1}{a + b \tan x} \text{ dx }\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×