Question

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

Answer 1

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

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

$=\int \frac{e^x}{e^x+1}dx$$-\int \frac{e^{-x}}{e^{-x}+1}dx$ [Multiplying the numerator and denominator of the second integral by $e^{-x}$]

$=\int \frac{d(e^x+1)}{e^x+1}dx$$+\int \frac{d(e^{-x}+1)}{e^{-x}+1}dx$

$=\log (1+e^x)+\log (1+e^{-x})+c$ [$c$ being integrating constant]

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

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

Comparing this with the given integral we get,

$f\left(x\right)=2\log (1+e^x)-x$