मराठी

∫ E X ( 1 + E X ) ( 2 + E X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

Let ex = t therefore ex dx = dt
`int    e^x/[( 1 + e^x)( 2 + e^x )]dx  = int dt/[( 1 + t)( 2 + t)]`

 = `int dt/( 1 + t) - int dt/( 2 + t)`

= log| 1 + t | - log| 2 + t | + c

= log `|( 1 + t )/( 2 + t )| + c`

= log `|( 1 + e^x )/( 2 + e^x )|`+ c

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.16 [पृष्ठ ९०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.16 | Q 14 | पृष्ठ ९०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

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

\[\int\sin x\sqrt{1 + \cos 2x} dx\]

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

`∫     cos ^4  2x   dx `


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

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

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

\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]

\[\int \sin^3 x \cos^5 x \text{ dx  }\]

Evaluate the following integrals:

\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

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

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

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]

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

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

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

\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]

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

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

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

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

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

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

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

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

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

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

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

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

\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 9}} \text{ dx}\]

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


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


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

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

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

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

Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×