A) \[\log ({{e}^{x}}-1)+c\]
B) \[-\log ({{e}^{x}}-1)+c\]
C) \[\log (1-{{e}^{x}})+c\]
D) None of these
Correct Answer: A
Solution :
\[\int{\frac{{{e}^{x}}}{{{e}^{x}}-1}}dx\] Let \[{{e}^{x}}-1=t\Rightarrow {{e}^{x}}dx=dt\] \[\therefore \] \[\int{\frac{{{e}^{x}}}{{{e}^{x}}-1}}dx=\int{\frac{1}{t}}dt\] \[=\log t+c\] \[=\log ({{e}^{x}}-1)+c\]
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