Double integral finding new limit $\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv$ I don't understand how to find new limits when I use substitution method.
I have this integral :
$$\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv=$$
$$t=u^2+4v, dt=2udu, \frac{dt}{2}=udu$$
The new limits for $du$ are $\int_{4v}^{1+4v}$ after I substitute $u=1,u=0$ for $t$.
But for some reason in my book they get $\int_{0}^{1}$, but I don't understand what about $4v$
Any ideas?
Any help will be appreciated, Thanks in advance!
EDIT (The full answer in the book): 
$$\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv=\int_1^2(\sqrt{u^2+4v}|_{u=0}^{u=1})dv=\int_1^2(\sqrt{4v+1}-2\sqrt{v})dv=\\\frac{2}{3}(\frac{1}{4}\sqrt{(4v+1)^3}-2\sqrt{v^3} |_{1}^{2}=\frac{35}{6}-\frac{5\sqrt{5}}{5}-\frac{8\sqrt{2}}{3}$$

In both approaches, first the inner integral is evaluated and then the outer. Since there is no dispute about the outer integral, we look at how the inner integrals are done.
Let look at your approach first. First you look at the inner integral and write
\begin{align}
\int_0^1 \frac{u}{\sqrt{u^2+4v}} du&=\frac12\int_{4v}^{1+4v} \frac{1}{\sqrt{t}} dt\\
&=\sqrt{1+4v}-2\sqrt{v}
\end{align}
Now let look at how your book handle the inner integral:
\begin{align}
\int_0^1 \frac{u}{\sqrt{u^2+4v}} du&=\sqrt{u^2+4v}\Big|_{u=0}^{u=1}\\
&=\sqrt{1+4v}-2\sqrt{v}
\end{align}
Therefore both are identical.
