<s>[INST] How to evaluate this definite integral? Here is the integral we have to evaluate:

$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$

After observing, i realise that i can't evaluate these integrals from area of circle, I say that $u=\sqrt{x^2+4}$. Then i can say $dx=\frac{udu}{\sqrt{u^2-4}}$. The first term would transform into:
$$\int_2^{2\sqrt{5}}\frac{u^2\,du}{\sqrt{u^2-4}}$$
Similarly i say $v=\sqrt{x^2-4}$. Then i can also say $dx=\frac{vdv}{\sqrt{v^2+4}}$ and the second term would transform into:
$$\int_0^4\frac{v^2\,dv}{\sqrt{v^2+4}}$$
But this integrals also doesn't seem easy to solve. It goes without saying  that after substitution, the bounds of integrals interchanged. Maybe it can be helpful to solve the problem. Thank you for your effort!
 [/INST] Hint -
1.) $\int \sqrt{x^2+a^2} = \frac 12 x \sqrt{x^2+a^2} - \frac{a^2}2 sinh^{-1}\frac xa + c$
Or 
$= \frac 12 x \sqrt{x^2+a^2} + \frac{a^2}2 \ln|x + \sqrt{x^2+a^2}+c$
2.) $\int \sqrt{x^2-a^2} = \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 cosh^{-1}\frac xa + c$
Or 
$= \frac 12 x \sqrt{x^2-a^2} - \frac{a^2}2 \ln|x + \sqrt{x^2-a^2} + c$
For more formulas See this link.
After applying these formulas fill limits.
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