Advertisements
Advertisements
Question
\[\int\frac{dx}{e^x + e^{- x}}\]
Sum
Advertisements
Solution
\[\int\frac{dx}{e^x + e^{- x}}\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[\text{ let } e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} \left( t \right) + c\]
\[ = \tan^{- 1} \left( e^x \right) + c\]
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
\[\int\frac{x^5 + x^{- 2} + 2}{x^2} dx\]
\[\int\frac{1}{1 - \cos 2x} dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int\frac{2x - 1}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
` ∫ cos mx cos nx dx `
\[\int\frac{e^x + 1}{e^x + x} dx\]
\[\int\frac{1}{ x \text{log x } \text{log }\left( \text{log x }\right)} dx\]
\[\int\frac{x}{\sqrt{x^2 + a^2} + \sqrt{x^2 - a^2}} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\ \int\ x \left( 1 - x \right)^{23} dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int\frac{e^{3x}}{4 e^{6x} - 9} dx\]
\[\int\frac{3 x^5}{1 + x^{12}} dx\]
\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]
\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]
\[\int x \text{ sin 2x dx }\]
\[\int \sec^{- 1} \sqrt{x}\ dx\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int\left( e^\text{log x} + \sin x \right) \text{ cos x dx }\]
\[\int e^x \sec x \left( 1 + \tan x \right) dx\]
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
Find \[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to
\[\int\frac{x^3}{x + 1}dx\] is equal to
\[\int\text{ cos x cos 2x cos 3x dx}\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int x \sin^5 x^2 \cos x^2 dx\]
\[\int \sec^4 x\ dx\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]
