\int {1}/{(e^{x}-1)^{2}} d x=

Correct Answer is : log[exex−1]−1ex−1+c

\log [ {e^{x}}/{e^{x}-1}]+\frac{1}{e^{x}-1}+c

\log [{e^{x}-1}/{e^{x}}]+\frac{1}{e^{x}-1}+c

\log[ {e^{x}}/{e^{x}-1}]-\frac{1}{e^{x}-1}+c

\log[ {e^{x}}/{e^{x}-1}]-\frac{1}{e^{x}+1}+c

Integral Calculus

Question : 68 of 73

Marks: +1, -0

∫ \;{1}/{(e^{x}-1)^{2}} d x=

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