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
Advertisements

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

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

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

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

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

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

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

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]

\[\int\frac{e^\sqrt{x} \cos \left( e^\sqrt{x} \right)}{\sqrt{x}} dx\]

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

\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]

` ∫    \sqrt{tan x}     sec^4  x   dx `


\[\int \sin^3 x \cos^6 x \text{ dx }\]

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

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

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

\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]

\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]

`int"x"^"n"."log"  "x"  "dx"`

 
` ∫  x tan ^2 x dx 

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

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

\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]

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

\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

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

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

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]


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

\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\] 

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

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

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

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

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

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×