Q:

How to combine these 2 notations to be as simple as possible?

Consider the following question:
Find the antiderivative of $f(x) = 6x \, (x^2+1)^5$. 
I have been using 2 notations and I would like to combine them. 

Notice that $[x^2 + 1]' = 2x$, so $6x \, dx = 3 d(x^2+1)$
Take $(x^2 + 1) = u$, then $6x$ is $3\,u'$.

I would like to combine these, but I'm wondering if it would still be correct:
Take $[x^2 +1] = u$, then $6x = 3 \, du$. This would save me a lot of time on tests, and also would make it easier, however with calculus I am never sure if my notation is correct.
Is this correct, and if not; what would the optimal way to combine these methods, to be as concise as possible, be? 
By the way, this would be how I would do the problem:
Take $[x^2 +1] = u$, then $6x = 3 \, du$. 
$6x(x^2+1)^5 = 3du \cdot u^5 = 3u^5 du = d \dfrac{1}{2}u^6 = d \dfrac{1}{2}(x^2+1)^6$.
$F(x) = \dfrac{1}{2} (x^2+1)^6 + c$.

A:

Yes, that will work: $$u = x^2 + 1 \;\implies\; du = 2x\,dx,\;\text{so}\; 6x\,dx = 3 du$$
and the resulting integral will be $$\int 6x(x^2 + 1)^5 \,dx = \int u^5 \,(3 du) = 3\int u^5\,du$$
Now we just integrate with respect to $u$:
$$3\int u^5\,du = 3\left[\frac{u^6}{6}\right] + C$$
Back substituting gives us $$\frac 12(u^6) + C = \frac 12(x^2 + 1)^6 + C$$

