Find the integral of this equation $\int \frac {1-e^x}{1+e^x}dx $

Question

Find: $$\int \frac {1-e^x}{1+e^x}dx $$ My solution: $$u = 1 + e^x,\;e^x = u - 1$$ $$\frac{du}{dx}=e^x,\;du=e^x dx,\;dx=\frac{du}{e^x}$$ $$\int \frac {1-u + 1}{u}\;\frac{du}{u-1} $$ $$\int \frac {2-u}{u}\;\frac{du}{u-1} $$ $$\int \frac {2-u}{u^2-u}\;du $$ $$\int \frac {2}{u^2-u}\;du\; -\;\int \frac {1}{u-1}\;du $$ Can I do this? $$\frac {2\ln(u^2-u)}{2u-1}\;-\ln(u-1)$$ $$\frac {2\ln((1+e^x)^2-(1+e^x))}{2(1+e^x)-1}\;-\;\ln(e^x)$$ $$\frac {2\ln(e^x+e^{2x})}{1+2e^x}\;-\;\ln(e^x)$$

Answer 1

I think an easier approach would be to multiply by $e^{-\frac{x}{2}}$ to obtain $$ \frac{1-e^x}{1+e^x}=\frac{e^{-\frac{x}{2}}-e^{\frac{x}{2}}}{e^{-\frac{x}{2}}+e^{\frac{x}{2}}}$$ and then make the substitution $u=e^{-\frac{x}{2}}+e^{\frac{x}{2}}$.

Alternately, write $$ \frac{1-e^x}{1+e^x}=1-2\frac{e^x}{1+e^x}$$ and in the second term use the substitution $u=1+e^x$.

Answer 2

Your Error: $$\int \frac{2}{u^2-u}=\int \frac{2(u-(u-1))}{u(u-1)}=2\int \frac{1}{u-1}-\frac{1}{u}$$ $$=2\ln\frac{u-1}{u}$$

Answer 3

$\displaystyle \int \frac{1-e^x}{1+e^x}dx=\int \tanh(\frac x2)dx=2\ln(\cosh(\frac x2))$

Answer 4

Easier way:

$$\int \frac{1-e^x}{1+e^x}dx=\int \frac{1+e^x-2e^x}{1+e^x}dx=\int 1dx-2\int\frac{de^x}{1+e^x}=x-2\ln{(1+e^x)}+C.$$