How to calculаte $\int_{\ln\sqrt{8}} ^{\ln\sqrt{24}}\sqrt{1+e^{2x}} \, dx$ I need help with calculating this integral:
$$\int_{\ln\sqrt{8}} ^{\ln\sqrt{24}}\sqrt{1+e^{2x}} \, dx$$
I got the answer using five replacements and I think my long solution is not optimal. Please give me advice what methods and how should I use.
 A: HINT
$$1 + e^{2x} = z ~~~~~~~~~~~~~ \text{d}z = 2 e^{2x} \,\text{d}x \to \text{d}z= 2(z-1) \,\text{d}x ~~~~~~~~~~~ \text{d}x = \frac{\text{d}z}{2(z-1)}$$
It's thence straightforward.
Some more details
Done that, and you get (forget the limits of the integrand, treat it as indefinite and then at the end go back to $x$ with those limits, in order to avoid complications and errors):
$$\frac{1}{2}\int \frac{\sqrt{z}\ \text{d}z}{z-1}$$
This gives you the result
$$2 \left(\sqrt{z}-\tanh^{-1}\left(\sqrt{z}\right)\right)$$
Obtained by using another substitution like $p = \sqrt{z}$ then partial fractions.
Then you can go back to $x$ and substitute the numerical limits.
If you do everything correctly, you will get as a final result:
$$2+\tanh^{-1}(3)-\tanh^{-1}(5)$$
A: The inverse function of $\sqrt{1+e^{2x}}$ is $\frac{1}{2}\log(x^2-1)$, so
$$ \int_{\log\sqrt{8}}^{\log\sqrt{24}}\sqrt{1+e^{2x}}\,dx + \int_{3}^{5}\frac{1}{2}\log(x^2-1)\,dx = 5\log\sqrt{24}-3\log\sqrt{8}$$
and, due to $\log(x^2-1)=\log(x+1)+\log(x-1)$,
$$\int_{3}^{5}\log(x^2-1)\,dx = \int_{2}^{4}\log x\,dx + \int_{4}^{6}\log x\,dx = \int_{2}^{6}\log x\,dx = \left[x\log x-x\right]_{2}^{6}$$
leads to
$$ \int_{\log\sqrt{8}}^{\log\sqrt{24}}\sqrt{1+e^{2x}}\,dx = 2+\text{arctanh}(3)-\text{arctanh}(5)=2+\log\frac{2}{\sqrt{3}}.$$
A: $$\int_{\ln\sqrt{8}} ^{\ln\sqrt{24}}\sqrt{1+e^{2x}} \, dx$$
$$1+e^{2x}=t \Rightarrow e^{2x}=t-1 \Rightarrow 2x=\ln(t-1) \Rightarrow x={{\ln(t-1)}\over 2}=\ln{\sqrt{t-1}} $$ $$dx={1\over 2} \cdot {dt\over {t-1}}$$
$$x=\ln(\sqrt8) \Rightarrow t=9;\ x=\ln(\sqrt24) \Rightarrow t=25$$
$$\int_{9} ^{25}{\sqrt{t}dt\over 2({t-1})}=\int_{9} ^{25}{\sqrt{t}dt\over 2({\sqrt t-1})({\sqrt t+1})}=\int_{9} ^{25}{(\sqrt{t}+1-1)dt\over 2({\sqrt t-1})({\sqrt t+1})}=$$
$$=\int_{9} ^{25}{dt\over 2({\sqrt t-1})}-\int_{9} ^{25}{dt\over 2({t-1})}$$
$$integral_1=\int_{9} ^{25}{dt\over 2({\sqrt t-1})}=\int_{9} ^{25}{\sqrt tdt\over 2\sqrt t({\sqrt t-1})}=\int_{9} ^{25}{\sqrt td(\sqrt t)\over {\sqrt t-1}}=\int_{9} ^{25}{(\sqrt t-1+1)d(\sqrt t)\over {\sqrt t-1}}=$$
$$=\int_{9} ^{25}{d(\sqrt t)}+\int_{9} ^{25}{d(\sqrt t)\over {\sqrt t-1}}=\sqrt t \biggr |_9^{25}+\ln(\sqrt t - 1)\biggr |_9^{25}=5-3+\ln4-\ln2=2+\ln2$$
$$integral_2=-\int_{9} ^{25}{dt\over 2({t-1})}={1\over 2}\ln(t-1)\biggr |_9^{25}=-0.5\ln24+0.5\ln8=0.5\ln({1\over 3})$$
Result is

$$2+\ln2+0.5\ln({1\over 3})\approx 2.14$$

