Question

lf $\int \frac{e^x-1}{e^x+1}dx=f\left(x\right)+c,$ then $f\left(x\right)=$

• A. $2\log (e^x+1)$
• B. $\log (e^{2x}-1)$
• C. $2\log (e^x+1)-x$
• D. $\log (e^{2x}+1)$

Answer 1

Answer: (C)

$2\log (e^x+1)-x$

Substitute $e^x=t$.

$e^x.dx=dt$

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

$\int \left(\frac{t-1}{t+1}\right)\frac{dt}{t}$

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

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

$=2\log (t+1)-\log t+c$

$=2\log (e^x+1)-\log e^x+c$

$=2\log (e^x+1)-x+c$