Is $\int (x-1)dx$ and $\int(x-1)d(x-1)$ same? Just confuse whether $$\int (x-1)\,dx\quad\text{and}\quad\int (x-1)\,d(x-1)$$ have same result? Many people tell me they are same (even math PhD), their reason is $dx=d(x-1)$, so answer must be same. But personally, I think the first one, answer is $x^2/2-x+C$, and second one is $(x-1)^2/2$. And I think the $\int(x-1)\,d(x-1)$ is to integral the area bases on $y$-axis. If I am wrong, tell me why and how I wrong?
Besides, any suggest when we deal with the situation such as $d\sin(x),\ de^x\ \text{and}\ d\ln(x)$.
 A: It might depend on what you mean by $d(x-1)$ in the integral.
You have 
$$
\int x - 1\; dx = \frac{1}{2}x^2 - x + c.
$$
Now the integral
$$
\int x-1\; d(x-1)
$$
is usually understood to be the integral
$$
\int u\; du
$$
where $u = x-1$. Here indeed $du = dx$ and so the integral is
$$
\int u\; du = \frac{1}{2}u^2 + d = \frac{1}{2}(x-1)^2 + d = \frac{1}{2}x^2 - x + \frac{1}{2} + d.
$$
You can "absorb" the $\frac{1}{2}$ into the constant, and so the integrals are indeed the same.
A: $$\int (x-1)\ \text{d}x = \frac{x^2}{2} - x + C$$
$$\int (x-1)\ \text{d}(x-1) \equiv \int z\ \text{d}z = \frac{z^2}{2} + C = \frac{(x-1)^2}{2} + C$$
Back to $x$:
$$ = \frac{x^2}{2} + \frac{1}{2} - x + C = \frac{x^2}{2} - x + C'$$
Thence they are indeed the same modulo the constants C and C'.
A: I notice that in asking the question you included a constant $C$ in your first integral, but omitted to do so in your second.  That constant — which ought to be in both expressions — is the clue to explaining why these two formulas are in fact equivalent.
I think it is important, first, to realize that when we write an indefinite integral like
$$\int {x-1} \space  dx=\frac{1}{2}x^2 -x + C$$
the right-hand-side denotes not a single function, but an entire family of functions that differ from one another only be a constant.  Put another way:  For any constant $C$, the function $\frac{1}{2}x^2 -x + C$ is an antiderivative of $x-1$.
Now consider your second integral.  You wrote:
$$\int {x-1} \space d(x-1)=\frac{1}{2}(x-1)^2$$
If you multiply out the right-hand side, you get $\frac{1}{2}(x^2-2x+1)=\frac{1}{2}x^2-x+\frac{1}{2}$.  Compare this with the result above it, and you see that it is exactly the same thing, with $C=\frac{1}{2}$.  So in fact $\frac{1}{2}(x-1)^2$ is a member of the family of functions denoted by $\frac{1}{2}x^2 -x + C$.
ETA:  Regarding other differentials:  It's not clear to me what exactly you are asking, but here are a few equivalences that might help:
$$d(e^x)=e^x\space dx$$
$$d(\sin x) = \cos x \space dx$$
$$d(\ln x) = \frac{1}{x} \space dx$$
