Evaluate : $\int \frac {e^{2x} -1}{e^x +1} dx$

Question

Evaluate : $$\int \dfrac {e^{2x} -1}{e^x +1} dx$$

My Attempt: $$\int \dfrac {e^{2x}-1}{e^x+1}dx\\ =\int \left(\dfrac {e^{2x}}{e^x +1} - \dfrac {1}{e^x+1}\right)dx\\ =\int \dfrac {e^{2x}}{e^x+1} dx - \int \dfrac {dx}{e^x+1}$$

Answer 1

I will continue from where you left $$=\int \dfrac {e^{2x}}{e^x+1} dx - \int \dfrac {dx}{e^x+1}$$ Now $$\int \dfrac {e^{2x}}{e^x+1} dx$$ Apply $u-$substitution $u=e^x+1$ and we get $$\int \dfrac {e^{2x}}{e^x+1} dx=\int\frac{u-1}{u}du=u-\ln|u|=e^x+1-\ln|e^x+1|$$ And $$ \int \dfrac {1}{e^x+1}dx=\int1-\dfrac {1}{e^x+1}dx=x-\ln|e^x+1|$$ $$\int \dfrac {e^{2x}}{e^x+1} dx - \int \dfrac {dx}{e^x+1}=e^x+1-\ln|e^x+1|-x+\ln|e^x+1|$$ $$=e^x-x+1+C$$ $$\int\frac{e^{2x}-1}{e^x+1}=e^x-x+1+C$$

Answer 2

$$\int\frac{e^{2x}-1}{e^{x}+1}\,dx =\int\left(e^x-\frac{e^x+1}{e^x+1}\right)\,dx =\int e^xdx-\int1 dx =e^x-x+c, \quad c\in\mathbb R$$

Answer 3

$$ \int \dfrac {e^{2x} -1}{e^x +1} dx = \int \dfrac {(e^{x} -1)(e^{x} + 1)}{e^x +1} dx=\int ({e^x -1}) dx=e^x - x +C $$