हिंदी

∫ E X + 1 E X + X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{Let I} = \int\frac{e^x + 1}{e^x + x}dx\]
\[\text{Putting }e^x + x = t\]
\[ \Rightarrow e^x + 1 = \frac{dt}{dx}\]
\[ \Rightarrow \left( e^x + 1 \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln }\left| t \right| + C\]
\[ = \text{ln }\left| e^x + x \right| + C \left[ \because t = e^x + x \right]\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.08 | Q 17 | पृष्ठ ४७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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


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

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

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


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

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

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

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

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

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

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

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

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


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


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

Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]

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

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

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

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

\[\int2 x^3 e^{x^2} dx\]

\[\int {cosec}^3 x\ dx\]

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

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

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

\[\int \sin^3 \sqrt{x}\ dx\]

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

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

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

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

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


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

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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×