Question

If $\int \frac{e^x-1}{e^x+1}dx=f\left(x\right)+C$, then f(x) is equal to

• A. $2\log \left|e^x+1\right|$
• B. $\log \left|e^{2x}-1\right|$
• C. $e^x-2\log \left|e^x+1\right|$
• D. $\log \left|e^{2x}+1\right|$

Answer 1

The correct option is C $e^x-2\log \left|e^x+1\right|$

$\int \frac{e^x-1}{e^x+1}dx$ $e^x=t$

$e^{x}dx=t$ $dx=\frac{dt}{t}$

$=\int \frac{t-1}{t+1}\frac{dt}{t}=\int \frac{t-1}{t(t+1)}ddt$

$=\int (1-\frac{2}{t+1})dt$

$=t-2\log |e^x+1|$

$=e^x-2\log |e^x+1|+C$