I think that for solving
$\int_0^\infty 1-tanh(x)dx$
I have to use the fact that $tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$, so that the integral becomes:
$2\int_0^\infty \frac{e^{-x}}{e^{x}+e^{-x}}dx$
At this point I was thinking to change variable. Maybe something like $u=e^{x}+e^{-x}$ or $u=e^{-x}$. But it did not work.
Can someone help, please?
HINT: Use the change of variable $x \to -\ln u$. Then you have $$-2\int \frac{1}{u} \frac{e^{\ln u}}{e^{-\ln u}+e^{\ln u}}$$ $$-2\int \frac{1}{u} \frac{u}{\frac{1}{u}+u}$$ $$-2\int \frac{u}{1+u^2}$$ I think you can figure it out from here...
$$ \int_{0}^{\infty} 1- \tanh x \, dx = \ln 2 $$
You are one the right track by writing the problem in exponential form.
Substitution
We have $$ \tanh x = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $$ Use the substitution $$ \color{blue}{u = \cosh x}, \qquad \color{blue}{du = \sinh x \, dx} $$ The integral is now $$ \int \tanh x \, dx = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx = \int \frac{du}{u} = \ln u $$ The full integral is $$ \int 1- \tanh x \, dx = x - \ln \cosh x $$
Take the limits $$ \lim_{x\to \infty} \left( x - \ln \cosh x \right) = \lim_{x\to\infty} \left( x - \ln \frac{e^{x}}{2} \right) = \lim_{x\to\infty} \left( x - \ln e^{x} + \ln 2 \right) = \ln 2 $$ The limit $x\to 0$ is trivial. The final answer is $$ \boxed{ \int_{0}^{\infty} 1- \tanh x \, dx = \ln 2 } $$
Hint:
$$ \int (1-\tanh x)dx = \int dx-\int \tanh x dx $$ and $$ \int \tanh x dx=\int \frac{\sinh x}{\cosh x} dx=\ln(\cosh(x)) +C $$ is a standard integral, easy calculated with the substitution $$ \cosh x= u \quad \rightarrow \quad du=\sinh x\, dx$$
First I tried simplifying the integrand by multiplying and dividing by $e^x$. This gave
$2 \int_0^\infty \dfrac{1}{e^{2x}+1} \ dx$.
It would be nice to use u-substitution with $u=e^x$ but it doesn't work. But notice that you can multiply and divide by $e^x$ again:
$2 \int_0^\infty \dfrac{e^x}{(e^{2x}+1)e^x} \ dx$
Now let $u=e^x$. The integral becomes
$2 \int_1^\infty \dfrac{1}{(u^2+1)u} \ du$.
Then you can use partial fractions by writing $\dfrac{1}{(u^2+1)u}$ in the form $\dfrac{A}{u}+\dfrac{Bx+C}{u^2+1}$, solving for $A$ and $B$, and then integrating. The main steps can be found here: https://www.mathway.com/popular-problems/Calculus/564881
