Evaluate $\int_0^1\frac{\mathrm{e}^{12x}-\mathrm{e}^{-12x}}{\mathrm{e}^{12x}+\mathrm{e}^{-12x}}\,\mathrm{d}x$ My work so far
$${\displaystyle\int_0^1}\dfrac{\mathrm{e}^{12x}-\mathrm{e}^{-12x}}{\mathrm{e}^{12x}+\mathrm{e}^{-12x}}\,\mathrm{d}x$$
using substitution with $u=\mathrm{e}^{12x}+\mathrm{e}^{-12x}$
$$={\displaystyle\int}\dfrac{1}{12u}\,\mathrm{d}u$$
$$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{12}}}{\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
$${\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u$$
which is the standard integral:
$$=\ln\left(u\right)$$
$$=\dfrac{\ln\left(u\right)}{12}$$
and using substitution with $u=\mathrm{e}^{12x}+\mathrm{e}^{-12x}$
$$=\dfrac{\ln\left(\mathrm{e}^{12x}+\mathrm{e}^{-12x}\right)}{12}+C$$
$$=\dfrac{\ln\left(\mathrm{e}^{-24}\left(\mathrm{e}^{24}+1\right)\right)}{12}-\dfrac{\ln\left(2\right)}{12}+1$$
which is rewritten as
$$\dfrac{\ln\left(\mathrm{e}^{-24}\left(\mathrm{e}^{24}+1\right)\right)-\ln\left(2\right)+12}{12}$$
Is my work correct so far? Also, I thought I was done at the last step, but I noticed this integral can be approximated to $0.9422377349564838$. How would I go about doing this part?
 A: $$\int_0^1\frac{e^{12x}-e^{-12x}}{e^{12x}+e^{-12x}}\ dx$$
$$=\frac{1}{12}\int_0^1\frac{d(e^{12x}+e^{-12x})}{e^{12x}+e^{-12x}}$$
$$=\frac{1}{12}[\ln(e^{12x}+e^{-12x})]_0^1$$
$$=\frac{1}{12}[\ln(e^{12}+e^{-12})-\ln(2)]$$
A: If you want to approximate this numerically, we'll have to resort to series
$$\frac{\log \cosh(12)}{12} = \frac{\log\left(\frac{e^{12}+e^{-12}}{2e^{12}}\right)+12}{12} \approx \frac{\log(1-\frac{1}{2})}{12}+1 \approx 1 - \frac{1}{24} - \frac{1}{96}= \frac{91}{96} \sim 0.948$$
You won't need to consider the $e^{-12}$ until you want around $8$ figures of precision.
A: We have
$$\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=\frac{\sinh(ax)}{\cosh(ax)}=\tanh(ax)$$
Then the integral becomes
$$\int_0^1 \frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}dx=\int_0^1\tanh(ax)dx=\left. \frac{\ln(\cosh(ax))}{a}\right|_{x=0}^{x=1}=\frac{\ln(\cosh(a))}{a}$$
With $a=12$ we get
$$\int_0^1 \frac{e^{12x}-e^{-12x}}{e^{12x}+e^{-12x}}dx=\frac{\ln(\cosh(12))}{12}$$
A: Apply the limits to
$$\dfrac{\ln\left(\mathrm{e}^{12x}+\mathrm{e}^{-12x}\right)}{12}\bigg|_0^1= \frac1{12}\ln\frac{e^{12}+e^{-12}}{2}=\frac1{12}\ln \cosh12
$$
