Advertisements
Advertisements
प्रश्न
\[\int\frac{dx}{e^x + e^{- x}}\]
बेरीज
Advertisements
उत्तर
\[\int\frac{dx}{e^x + e^{- x}}\]
\[ = \int\frac{dx}{e^x + \frac{1}{e^x}}\]
\[ = \int\frac{e^x dx}{e^{2x} + 1}\]
\[\text{ let } e^x = t\]
\[ \Rightarrow e^x dx = dt\]
\[Now, \int\frac{e^x dx}{e^{2x} + 1}\]
\[ = \int\frac{dt}{1 + t^2}\]
\[ = \tan^{- 1} \left( t \right) + c\]
\[ = \tan^{- 1} \left( e^x \right) + c\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
\[\int\frac{1}{\sqrt{1 + \cos x}} dx\]
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
` = ∫ root (3){ cos^2 x} sin x dx `
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
\[\int \tan^3 \text{2x sec 2x dx}\]
\[\int x^2 \sqrt{x + 2} \text{ dx }\]
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
` ∫ \sqrt{tan x} sec^4 x dx `
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
`int 1/(cos x - sin x)dx`
\[\int x e^{2x} \text{ dx }\]
\[\int x\left( \frac{\sec 2x - 1}{\sec 2x + 1} \right) dx\]
\[\int e^x \left( \frac{x - 1}{2 x^2} \right) dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int(2x + 5)\sqrt{10 - 4x - 3 x^2}dx\]
\[\int\frac{2x + 1}{\left( x + 1 \right) \left( x - 2 \right)} dx\]
\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]
\[\int\frac{x^4}{\left( x - 1 \right) \left( x^2 + 1 \right)} dx\]
\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]
If `int(2x^(1/2))/(x^2) dx = k . 2^(1/x) + C`, then k is equal to ______.
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int \cos^3 (3x)\ dx\]
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
\[\int\frac{1}{4 x^2 + 4x + 5} dx\]
\[\int\frac{1}{1 - 2 \sin x} \text{ dx }\]
\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]
\[\int\frac{\log \left( 1 - x \right)}{x^2} \text{ dx}\]
\[\int\sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \text{ dx}\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]
