Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{Let I} = \int\frac{1}{e^x + 1}dx\]
\[ = \int\frac{e^{- x}}{1 + e^{- x}}dx\]
\[Putting\ e^{- x} = t\]
\[ \Rightarrow - e^{- x} = \frac{dt}{dx}\]
\[ \Rightarrow e^{- x} dx = - dt\]
\[ \therefore I = \int\frac{- 1}{1 + t}dt\]
\[ = - \text{ln} \left| 1 + t \right| + C\]
\[ = - \text{ln } \left| 1 + e^{- x} \right| + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integrals:
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: \[\int\left( 1 - x \right)\sqrt{x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
