evaluate the integral: $\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$ $$\int_0 ^\sqrt5  \frac{4x}{\sqrt{x^2+4}}dx$$
I got $4(5^\frac{1}{4})$ but I'm not sure if thats right

Your computation is incorrect.
The substitution $u=x^2+4$ gives $du = 2x\,dx$. When $x=0$, we have $u=4$; when $x=\sqrt{5}$, we get $u=9$. So
$$\begin{align*}
\int_0^{\sqrt{5}}\frac{4x}{\sqrt{x^2+4}}\,dx &= 2\int_0^{\sqrt{5}}\frac{2x\,dx}{\sqrt{x^2+4}}\\
&= 2\int_4^9\frac{du}{\sqrt{u}}\\
&= 2 \int_4^9 u^{-1/2}\,du\\
&= 4u^{1/2}\Bigm|_4^9\\
&= 4(9^{1/2} - 4^{1/2})\\
&= 4(3-2)\\
&= 4.
\end{align*}$$
