d x 1-e^x is equal to :

MCQ

$\int \frac{d x}{1-e^x}$ is equal to :

✓
$-\log \left(1-e^{-x}\right)+C$
B
$\log \left(1+e^{x}\right)+ C$
C
$\log \left(1+e^{-x}\right)+ C$
D
$\log \left(\frac{1+e^x}{e^{-x}}\right)+ C$

✓

Answer

Correct option: A.

$-\log \left(1-e^{-x}\right)+C$

$\int \frac{d x}{1-e^x}$
Let $1-e^x=t$
$\therefore-e^x d x =d t$
$d x =\frac{d t}{-e^x}=\frac{d t}{t-1}$
$\int \frac{d t}{t(t-1)}$
$\Rightarrow \int\left(\frac{1}{t-1}-\frac{1}{t}\right) d t =\log (t-1)-\log t+C$
$=\log \frac{(t-1)}{t}+C$
$=\log \left(\frac{1-e^x-1}{1-e^x}\right)+C$
$=\log \left(\frac{e^x}{e^x-1}\right)=-\log \left(\frac{e^x-1}{e^x}\right)+C$
$=-\log \left(1-e^{-x}\right)+C$
Hence option $(A)$ is correct.

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