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!
 A: Let $f(x)=\sqrt{x^2+4}$ and $f^{-1}(x)=\sqrt{x^2-4}$. Note that
$$f(4)=2\sqrt{5} \\
f(0)=2
$$
We know that the integral of an inverse function can be calculated with

$$\int_{f(k)}^{f(l)}f^{-1}(x)\,dx=lf(l)-kf(k)-\int_k^lf(x)\,dx$$

So the integral
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
can be written as
$$\int_0^4\sqrt{x^2+4}\,dx+4\cdot2\sqrt{5}-2\cdot0-\int_0^4\sqrt{x^2+4}\,dx$$
Finally the result is
$$ \bbox[5px,border:2px solid red]
{
8\sqrt{5}
}
$$
A: HINT: for the first integral use Integration by parts, $$u'=1,v=\sqrt{x^2+4}$$
for the second integral use $$x=\cosh(t)$$
A: You have to use a trigonometric substitution in this case. Let me demonstrate it for the first integral.
let $x= 2\tan u$ and $dx= 2\sec^2 u du$. Applying this substitution on the integral $$\int_0^4 \root \of{x^2+4} dx$$
yields
$$\int_0^{\arctan 0.5}4sec^3u du$$
You can solve the integral above by using integration by parts
The same idea applies to your second integral but the use of $x = 2\sec u$ as a subsitution would be more appropiate.
A: 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.
A: HINT...for the first one substitute $x=2\sinh \theta$ and for the second substitute $x=2\cosh \theta$
