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

My Attempt:
$$\int \dfrac {e^x -1}{e^x+1}dx=(e^x -1) \int \dfrac {1}{e^x +1} dx$$
Let $e^x =t$
$$e^x\cdot dx=dt$$
Then,
$$\int \dfrac {e^x -1}{e^x+1}dx=(e^x -1) \int \dfrac {1}{t(t+1)} dt$$
 A: Here is another hint :
$$I=\int \frac {e^x -1}{e^x+1}dx=\int \frac {e^{-\frac x 2}(e^x -1)}{e^{-\frac x 2}(e^x+1)}dx$$
$$I=\int \frac {\sinh(\frac x 2)}{\cosh(\frac x 2)}dx$$
$$I=2\int \frac {du} u=2\ln(u)+K $$
$$\boxed{I=2\ln(\cosh(\frac x 2))+K}$$
A: HINT: $$\int \dfrac {e^x -1}{e^x+1}dx=\int\left(1-\frac2{e^x+1}\right)\,dx=\int\,dx-2\int\frac{dx}{e^x+1}$$
A: we have $$\frac{e^x-1}{e^x+1}=\frac{e^x+1-2}{e^x+1}=1-\frac{2}{e^x+1}$$ and Substitute $$e^x=t$$
the result should be $$2 \log \left(e^x+1\right)-x+C$$
A: Using OP's approach:
$$\int\frac{e^x-1}{e^x+1}dx$$
Let $e^x=t$ and so $e^xdx=dt$. The new integral can then be written:
$$\int\frac{t-1}{t+1}\frac{1}{t}dt$$
Notice that $\frac{t+1}{t-1}=\frac{2}{t+1}+\frac{-1}{t}$, so our integral is equal to:
$$\int\frac{2}{t+1}-\frac{1}{t}dt=\int\frac{2}{t+1}dt-\int\frac{1}{t}dt=2\ln(t+1)-\ln(t)+C$$
Remembering that $t=e^x$ we finally get:
$$2\ln(e^x+1)-\ln(e^x)+C=2\ln(e^x+1)-x+C$$
A: Hint:
Let $e^x+1=t,\implies e^x\ dx=dt\implies dx=\dfrac{dt}{t-1}$
$$\implies\int\dfrac{e^x-1}{e^x+1}\ dx=\int\dfrac{t-2}{t(t-1)}dt$$
Now $\dfrac{t-2}{t(t-1)}=\dfrac1{t-1}-2\cdot\dfrac{t-(t-1)}{t(t-1)}=\dfrac1{t-1}-2\left(\dfrac1{t-1}-\dfrac1t\right)=\dfrac2t-\dfrac1{t-1}$
