हिंदी

∫ X + 3 ( X + 4 ) 2 E X D X = - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • \[\frac{e^x}{x + 4} + C\]

  • \[\frac{e^x}{x + 3} + C\]

  • \[\frac{1}{\left( x + 4 \right)^2} + C\]

  • \[\frac{e^x}{\left( x + 4 \right)^2} + C\]

MCQ

उत्तर

\[\frac{e^x}{x + 4} + C\]

 

\[\text{Let }I = \int\frac{\left( x + 3 \right)}{\left( x + 4 \right)^2} e^x dx\]
\[ \Rightarrow \int\left[ \frac{x + 4 - 1}{\left( x + 4 \right)^2} \right] e^x dx\]
\[ \Rightarrow \int\left[ \frac{1}{\left( x + 4 \right)} - \frac{1}{\left( x + 4 \right)^2} \right] e^x dx\]
\[\text{As, we know that }\int e^x \left\{ f\left( x \right) + f'\left( x \right) \right\} dx = e^x f\left( x \right) + C\]
\[ \therefore I = \frac{e^x}{x + 4} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - MCQ [पृष्ठ २०१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
MCQ | Q 18 | पृष्ठ २०१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

Integrate the following integrals:

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

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

\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]

\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]

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

 ` ∫   1 /{x^{1/3} ( x^{1/3} -1)}   ` dx


\[\int \cos^7 x \text{ dx  } \]

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

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

 


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

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

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

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

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

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

\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]

\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{  dx }\]

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

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

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

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

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


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

\[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\]

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

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

\[\int \sec^4 x\ dx\]


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


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

\[\int\frac{\log x}{x^3} \text{ dx }\]

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

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

\[\int\frac{\cos^7 x}{\sin x} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×