Evaluate this integral : $\int \frac{e^{2x}-1}{e^{2x}+1}dx$ $$\int \frac{e^{2x}-1}{e^{2x}+1}dx$$
I had this question for today's test and still can't find out how to do it. Any hint is appreciated.
Thanks
 A: Hint : $$\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=\frac{\sinh(x)}{\cosh(x)}=\frac{u'}{u}$$ with $u=\cosh(x)$.
A: $$I=\int \frac{e^{2x}-1}{e^{2x}+1}dx$$
$$I=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}dx$$
$$e^{x}+e^{-x}=t$$
$$\implies \{e^{x}-e^{-x}\}dx=dt$$
$$I=\int\frac{dt}{t}$$
A: Set $e^x=y\implies x=\ln y,dx=\dfrac{dy}y$
$$\int\dfrac{e^{2x}-1}{e^{2x}+1}dx=\dfrac{y^2-1}{(y^2+1)y}\ dy$$
Use Partial Fraction Decomposition
$$\dfrac{y^2-1}{(y^2+1)y}=\dfrac{Ay+B}{y^2+1}+\dfrac Cy$$
$$\implies y^2-1=(Ay+B)y+C(y^2+1)=y^2(A+C)+By+C$$
$$\implies C=-1,A+C=1, B=0$$
A: Hint
$$\frac{e^{2x}-e^{2x}+e^{2x}-1}{e^{2x}+1}=\frac{2e^{2x}-e^{2x}-1}{e^{2x}+1}=-1+\frac{2e^{2x}}{e^{2x}+1}$$
A: The other two answers are a bit more elegant than mine, but my thinking was, quick and dirty,
$$\int\frac{e^{2x}-1}{e^{2x}+1}dx = \int\frac{e^{2x}}{e^{2x}+1}dx - \int\frac{1} {e^{2x}+1}dx = \int\frac{2e^{2x}}{2(e^{2x}+1)}dx + \int\frac{-2e^{-2x}}{2(1+e^{-2x})}dx = \frac{1}{2}(\ln(e^{2x}+1) + \ln(e^{-2x}+1)) + C = \frac{1}{2}\ln(2+e^{2x}+e^{-2x})+C$$
