Advertisements
Advertisements
प्रश्न
\[\int\frac{1}{e^x + e^{- x}} dx\]
बेरीज
उत्तर
\[\text{ Let I }= \int\frac{1}{e^x + e^{- x}}\text{ dx }\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{e^x dx}{\left( e^x \right)^2 + 1}\]
\[\text{ Putting e}^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + 1}\]
\[ = \tan^{- 1} t + C ..............\left( \because \int\frac{dt}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = \tan^{- 1} e^x + C............. \left( \because t = e^x \right)\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{e^x dx}{\left( e^x \right)^2 + 1}\]
\[\text{ Putting e}^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[ \therefore I = \int\frac{dt}{t^2 + 1}\]
\[ = \tan^{- 1} t + C ..............\left( \because \int\frac{dt}{a^2 + x^2} = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right)\]
\[ = \tan^{- 1} e^x + C............. \left( \because t = e^x \right)\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]
\[\int\left( x^e + e^x + e^e \right) dx\]
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
\[\int \sin^4 x \cos^3 x \text{ dx }\]
\[\int \sin^3 x \cos^6 x \text{ dx }\]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]
\[\int\frac{x^3}{x^4 + x^2 + 1}dx\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]
\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int x \cos^3 x\ dx\]
\[\int e^x \left( \cos x - \sin x \right) dx\]
\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{1}{x \log x \left( 2 + \log x \right)} dx\]
\[\int\frac{1}{\left( x - 1 \right) \left( x + 1 \right) \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x + 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
Write a value of
\[\int e^{3 \text{ log x}} x^4\text{ dx}\]
\[\int\left( x - 1 \right) e^{- x} dx\] is equal to
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \cot^4 x\ dx\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\sqrt{a^2 + x^2} \text{ dx }\]
\[\int e^{2x} \left( \frac{1 + \sin 2x}{1 + \cos 2x} \right) dx\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int \left( e^x + 1 \right)^2 e^x dx\]
\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]
