Question: Prove that the area bounded by the curve $y=\tanh x$ and the straight line $y=1$, between $x=0$ and $x=\infty$ is $\ln 2$

$\int^\infty \_0 (1-\tanh x) \mathrm{dx}= \biggr[x-\ln (\cosh x)\biggr]^\infty\_0\\\infty - \ln(\cosh \infty)+1$

How do I get $\ln 2$ because the line and curve intersect at infinity?

Answer: 
You need to evaluate the limit
$$\lim\_{x\rightarrow \infty} (x - \ln(\cosh(x)))$$

Note that
$$e^{\ln(\cosh(x)) -x} = e^{-x} \cosh(x) = \frac{1}{2}(1 + e^{-2x})\rightarrow \frac{1}{2}$$

Therefore
$$\lim\_{x\rightarrow \infty} (x - \ln(\cosh(x))) = -\ln(1/2) = \ln(2)$$

Hence
$$\int\_0^\infty (1-\tanh(x))dx = \lim\_{y\rightarrow \infty} [x - \ln(\cosh(x))]\_{x=0}^{x=y} = \ln(2) $$
Well, when you get the antiderivative, you can put in the limits and get

$ \begin{array}{l}
\lim\limits \_{x\rightarrow \infty }[ x-ln( cosh( x))] =\lim\limits \_{x\rightarrow \infty }\left[\frac{( x+ln( cosh( x)))( x-ln( cosh( x)))}{( x+ln( cosh( x)))}\right] =\lim\limits \_{x\rightarrow \infty }\left[\frac{x^{2} -ln^{2}( cosh( x))}{x+ln( cosh( x))}\right]\\
=\lim\limits \_{x\rightarrow \infty }\left[\frac{x^{2}}{x+ln( cosh( x))} -\frac{ln^{2}( cosh( x))}{x+ln( cosh( x))}\right] =\lim\limits \_{x\rightarrow \infty }\left[\frac{2x}{1+tanh( x)} -\frac{2tanh( x) ln( cosh( x))}{1+tanh( x)}\right]\\
\\
\lim\limits \_{x\rightarrow \infty }[ x-ln( cosh( x))] =\lim\limits \_{x\rightarrow \infty }\left[ 2\frac{x-tanh( x) ln( cosh( x))}{1+tanh( x)}\right]\\
This\ means\ it\ equals\ itself,\ because\ tanh( \infty ) =1
\end{array}$

After that, I was stuck... But maybe this will help you; or someone else to answer your problem.

But I do know that

\lim\limits \_{x\rightarrow \infty }[ x-ln( cosh( x))] =ln( 2)

I just don't know how.