Why derivative of substituted part isn't taken in u subsitution calculus I am trying to understand u substitution unsuccessfully for sometime. For me to remember something i need to understand it to last bit...until then it's not proven in my mind and could mean anything else and building anything on top of that is nothing more than speculations. 
My problem as of now is that i don't get the process, basic idea, or even the process. Each time i have tried i have gotten stuck at part where U is just substituted with original value and it's said that yes this is it. Funny part it works too. 
So is there like something proof of it. How and why does it work?
thanks in advance. First simply describe the general process as i have forgotten, and then explain why especially the subsitution part. 
 A: Recall that $u$-substitution is really the inverse rule of the chain rule, just like integration by parts is the inverse rule of the product rule. The essence of the chain rule is that 
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x},$$
which is why we like to write derivatives as ratios - often, when they look like they cancel, they really "do cancel," so to speak. Notice that
$$ \int g(u(x))u'(x) \mathrm{d}x = \color{#F01C2C}{\int g(u(x)) \frac{\mathrm{d}u}{\mathrm{d}x}\mathrm{d}x = \int g(u) \mathrm{d}u} = \int g(u) \mathrm{d}u,$$
where I've notated the important equality in red. The step in red is visibly related to the chain rule: the part that looks like it cancels really does cancel.
You ask for a proof, so let's give a proof! Since integrals "undo" differentiation, and I've claimed that the u-substitution is just the chain rule in reverse (essentially), that provides the color of the proof. Consider the function 
$$ F(x) = \int_{0}^x g(t)\mathrm{d}t.$$
Consider the function $F(u(x))$ and differentiate it:
$$ \begin{align}
F(u(x))' &= F'(u(x)) u'(x) = \frac{\mathrm{d}F}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\\
&=\frac{\mathrm{d}}{\mathrm{d}u}\int_{0}^{u(x)} g(u(t))\mathrm{d}t \cdot u'(x)\\
&= g(u(x))u'(x).
\end{align}$$
The the second fundamental theorem of calculus says that 
$$\begin{align} 
\int_a^b g(u(x))u'(x)\mathrm{d}x &= F(u(b)) - F(u(a)) \\
&= \int_{a}^{b} g(u(t))u'(t)\mathrm{d}t \\
&=\int_{a}^{b}g(u(t))\frac{\mathrm{d}u}{\mathrm{d}t}\mathrm{d}t.
\end{align}$$
Of course, we also know that $\displaystyle F(u(b)) - F(u(a)) = \int_{u(a)}^{u(b)} g(t) \mathrm{d}t = \int_{u(a)}^{u(b)} g(u) \mathrm{d}u$.
Voila!
A: Using $u$ substitution is not so much a theorem but a technique. The thought process is basically "This integral isn't so simple, so let me substitute something for $u$ to make it integral I can recognize. However, because of the chainrule I must also account for $du$." 
For example, take $\displaystyle \int \frac{2x}{x^2 + 5} dx$. There's no well-known, formulaic anti-derivative for this function. However, if we use the substitution $u = x^2 + 5$ and $du = 2x$, we get $\displaystyle \int \frac{1}{u} du$ and, hey, now this is something we can work with. This has a known anti-derivative, so we can integrate now and get $\ln |u| + C$. Substituting $u$ back in, we get $\ln |x^2 + 5| + C$. To take it one step further, we can get rid of absolute value bars since $x^2 + 5  > 0 \forall x \in \mathbb{R}$
