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
 A: 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
$$\begin{align*}
\int_0^{\sqrt{5}}\frac{4x}{\sqrt{x^2+4}}\,dx &= 2\int_0^{\sqrt{5}}\frac{2x\,dx}{\sqrt{x^2+4}}\\
&= 2\int_4^9\frac{du}{\sqrt{u}}\\
&= 2 \int_4^9 u^{-1/2}\,du\\
&= 4u^{1/2}\Bigm|_4^9\\
&= 4(9^{1/2} - 4^{1/2})\\
&= 4(3-2)\\
&= 4.
\end{align*}$$
A: There are two approaches to the definite integral when you use substitution.
$1$.  Use substitution as usual, in this case $u=x^2+4$ (or maybe better $u^2=x^2+4$), to find that one indefinite integral is $4u^{1/2}$.  Then replace $u$ by $x^2+4$ to find that $4(x^2+4)^{1/2}$ is an indefinite integral of your original function, and "plug in" as usual.
$2$. Use substitution, but substitute for everything. So our original integral is 
$$\int_{x=0}^{\sqrt{5}}\frac{4x\,dx}{\sqrt{x^2+4}}.$$
Note that I put in the $x$ in the limits of integration to remind myself that there is an $x$ hidden there.
Substitute, remembering to substitute for everything. So we get
$$\int_{u=2}^3 2u^{-1/2}\,du.$$
Now the substitution has been fully done. Integrate. We get $4u^{1/2}$. Do the plugging in. We get $4$.
In general I prefer the second approach. Use one or the other, and not, as you did, a hybrid of both.
A: Since it's an elementary integral, I'd solve it directly:
$$I=4\int_{0}^{\sqrt{5}} \frac{x}{\sqrt{4+x^2}} dx=4\int_{0}^{\sqrt{5}} \left({\sqrt{4+x^2}}\right)' dx = 4 \sqrt{4+x^2}\Bigm|_0^{\sqrt{5}}=4.$$
Q.E.D.
