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Question

Question: \[\log(1 + e^{x}) - x + e^{- x} + c\]...

log⁡(1+ex)−x+e−x+c\log(1 + e^{x}) - x + e^{- x} + c

A

log⁡(1+ex)+x+e−x+c\log(1 + e^{x}) + x + e^{- x} + c

B

∫11−e2x 6mudx=\int_{}^{}{\frac{1}{\sqrt{1 - e^{2x}}}\mspace{6mu} dx =}

C

x−log⁡[1+1−e2x]+cx - \log\lbrack 1 + \sqrt{1 - e^{2x}}\rbrack + c

D

x+log⁡[1+1−e2x]+cx + \log\lbrack 1 + \sqrt{1 - e^{2x}}\rbrack + c

Answer

∫11−e2x 6mudx=\int_{}^{}{\frac{1}{\sqrt{1 - e^{2x}}}\mspace{6mu} dx =}

Explanation

Solution

Put ∫dxsin⁡x+3cos⁡x=\int_{}^{}\frac{dx}{\sin x + \sqrt{3}\cos x} =

then it reduces to

logtan(x2+π2)+c{logtan}\left( \frac{x}{2} + \frac{\pi}{2} \right) + c