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Question

Question: \(\int_{}^{}{\frac{dx}{e^{x} - 1} =}\)...

∫dxex−1=\int_{}^{}{\frac{dx}{e^{x} - 1} =}

A

ln⁡(1−e−x)+c\ln(1 - e^{- x}) + c

B

−ln⁡(1−e−x)+c- \ln(1 - e^{- x}) + c

C

ln⁡(ex−1)+c\ln(e^{x} - 1) + c

D

∫x2sec⁡x3 6mudx=\int_{}^{}{x^{2}\sec x^{3}\mspace{6mu} dx} =

Answer

ln⁡(1−e−x)+c\ln(1 - e^{- x}) + c

Explanation

Solution

Put 2(sec⁡x+tan⁡x)+c2(\sec x + \tan x) + c then

∫x51(tan⁡−1x+cot⁡−1x) 6mudx=\int_{}^{}{x^{51}(\tan^{- 1}x + \cot^{- 1}x)\mspace{6mu} dx =}

x5252(tan⁡−1x+cot⁡−1x)+c\frac{x^{52}}{52}(\tan^{- 1}x + \cot^{- 1}x) + c.