$u$-substitution into integral For example we want to integrate  following integral
$$\int_0^4 2x\cdot((9+x^2)^{1/2})\,dx$$
I have read that if we denote $u$ as  $u=9+x^2$ then  $du=2x*dx$  everything is clear here but then there was used such kind of method: the author simply  put $x=0$ and $x=4$ into the original $u$ definiton  got $u=9$ and $u=25$  and   finally wrote integral as $$\int_{u=9}^{u=25} u^{1/2}\,du.$$
My question is: is it correct? Or if I will use the same method to solve   similar integrals next time, will I be wrong or not? thanks

The author is correct. Let me try to explain why by using your example.
$$\int_{x=0}^{x=4} 2x(9+x^2)^{\frac{1}{2}} \text{d}x$$
If we let $u=9+x^2$ we get, as you said, $\text du = 2x \;\text dx$. So 
$$\int_{x=0}^{x=4} 2x(9+x^2)^{\frac{1}{2}} \text{d}x = \int_{x=0}^{x=4} u^{\frac{1}{2}} \;\text du.$$
$f(u)= u^{\frac{1}{2}}$ can easily be plotted in a coordinate system with a horizontal $u$-axis and a vertical $y$-axis. If you view the integral as the area under the curve, what does it mean to take the integral from $x=0$ to $x=4$? Remember that $x=0$ is not the same as $u=0$. So we translate from $x$ to $u$ like this:
$$ x= 0 \Rightarrow u = 9+0^2 = 0$$
$$ x= 4 \Rightarrow u = 9+4^2 = 25$$
And we get
$$\int_{x=0}^{x=4} u^{\frac{1}{2}} \;\text du =\int_{u=9}^{u=25} u^{\frac{1}{2}} \;\text du.$$
