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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, $$\int \dfrac {e^x -1}{e^x+1}dx=(e^x -1) \int \dfrac {1}{t(t+1)} dt$$

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    $\begingroup$ How did you pull out $e^{x}-1$ from the integral? $\endgroup$ – clark Jan 20 '18 at 16:22
  • $\begingroup$ @clark is correct in asking this. You cannot remove any function depending on x from under the integral sign. $\endgroup$ – Chickenmancer Jan 20 '18 at 16:29
  • $\begingroup$ Oh sorry. That's my mistake. I thought it to be constant. $\endgroup$ – pi-π Jan 20 '18 at 16:36
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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 $$2 \log \left(e^x+1\right)-x+C$$

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Here is another hint :

$$I=\int \frac {e^x -1}{e^x+1}dx=\int \frac {e^{-\frac x 2}(e^x -1)}{e^{-\frac x 2}(e^x+1)}dx$$ $$I=\int \frac {\sinh(\frac x 2)}{\cosh(\frac x 2)}dx$$ $$I=2\int \frac {du} u=2\ln(u)+K $$ $$\boxed{I=2\ln(\cosh(\frac x 2))+K}$$

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HINT: $$\int \dfrac {e^x -1}{e^x+1}dx=\int\left(1-\frac2{e^x+1}\right)\,dx=\int\,dx-2\int\frac{dx}{e^x+1}$$

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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:

$$2\ln(e^x+1)-\ln(e^x)+C=2\ln(e^x+1)-x+C$$

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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 $\dfrac{t-2}{t(t-1)}=\dfrac1{t-1}-2\cdot\dfrac{t-(t-1)}{t(t-1)}=\dfrac1{t-1}-2\left(\dfrac1{t-1}-\dfrac1t\right)=\dfrac2t-\dfrac1{t-1}$

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