Advertisements
Advertisements
प्रश्न
\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]
विकल्प
- \[\frac{- e^{- x}}{e^x + e^{- x}} + C\]
- \[- \frac{1}{e^x + e^{- x}} + C\]
- \[\frac{- 1}{\left( e^x + 1 \right)^2} + C\]
- \[\frac{1}{e^x - e^{- x}} + C\]
MCQ
उत्तर
\[\frac{- e^{- x}}{e^x + e^{- x}} + C\]
\[\text{Let }I = \int\frac{2 dx}{\left( e^x + e^{- x} \right)^2}\]
\[ = \int\frac{2 dx}{\left( e^x + \frac{1}{e^x} \right)^2}\]
\[ = 2\int\frac{e^{2x} dx}{\left( e^{2x} + 1 \right)^2}\]
\[\text{Let }e^{2x} + 1 = t\]
\[ \Rightarrow e^{2x} \cdot 2 dx = dt\]
\[ \Rightarrow e^{2x} \cdot dx = \frac{dt}{2}\]
\[ \therefore I = 2 \times \frac{1}{2}\int\frac{dt}{t^2}\]
\[ = - \frac{1}{t} + C\]
\[ = - \frac{1}{e^{2x} + 1} + C ...............\left( \because t = e^{2x} + 1 \right)\]
Dividing numerator and denominator by ex
\[\Rightarrow I = \frac{- \frac{1}{e^x}}{e^x + \frac{1}{e^x}}\]
\[ = \frac{- e^{- x}}{e^x + e^{- x}} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]
\[\int\text{sin mx }\text{cos nx dx m }\neq n\]
\[\int\frac{\cos^5 x}{\sin x} dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
` ∫ tan^5 x sec ^4 x dx `
` ∫ tan^5 x dx `
\[\int \cos^5 x \text{ dx }\]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]
\[\int\frac{1}{4 x^2 + 12x + 5} dx\]
\[\int\frac{x - 1}{3 x^2 - 4x + 3} dx\]
\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{ dx}\]
\[\int\frac{1}{5 + 4 \cos x} dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int \sin^{- 1} \sqrt{x} \text{ dx }\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]
\[\int\left( 2x - 5 \right) \sqrt{x^2 - 4x + 3} \text{ dx }\]
\[\int\frac{x^2 + x - 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \cos^3 (3x)\ dx\]
\[\int \cos^5 x\ dx\]
\[\int\sqrt{\frac{1 + x}{x}} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
