Integrals using Arctangens We want to find $\displaystyle \int\dfrac{12}{16x^2 +1}$
I rewrote it to the form $ 3 \cdot \dfrac{1}{u^2 + 1} \cdot u' $ where $u=4x$.
I found out that the correction sheet does the same thing, but their next step leaves my puzzled:
$$ F(x) = 3 \arctan (4x) + C$$
Where did the $u' = 4$ go to?
 A: As you noted, if $u = 4x$, then $du = 4dx$, yielding
$$\begin{split}
\int \frac{12dx}{16x^2+1} 
 &= 3 \int \frac{4dx}{(4x)^2+1} \\
 &= 3 \int \frac{u'dx}{(u)^2+1} \\
 &= 3 \int \frac{du}{u^2+1} \\
 &= 3 \arctan(u) + C \\
 &= 3 \arctan(4x)+C
\end{split}
$$
A: It went to the composite function $arctan \circ u$ when it obtained by integrating $arctan'(u(x))u'(x)$.
A: If substitution is to be done successfully, it is a good idea to use standard notation, which was designed to make the work mechanical. The OP starts with "We want to find $\int \frac{12}{16x^2+1}$." That is not notationally correct. The use of incorrect notation may be part of the reason you find the subsitution problematic.
What we want to find is
$$\int \frac{12}{16x^2+1}\,dx.$$
Now we will go through the substitution process mechanically. Let $u=4x$.
We need to replace $x$ everywhere by the appropriate expression that involves $u$. From $u=4x$, we obtain $\frac{du}{dx}=4$. Write this as $du=4\,dx$, and then as 
$$dx=\frac{1}{4}\,du.$$
Now do the substitution. We must replace the $dx$ by the appropriate expression that only inovlves $u$. We get
$$\int \frac{12}{u^2+1}\cdot\frac{1}{4}\,du.$$
There is a bit of numerical cancellation. We get
$$\int\frac{3}{u^2+1}\,du.$$
This is a familiar integral, and the result is 
$$3\arctan u+C.$$
Finally, replace the $u$ by $4x$. 
Remark: Leaving out the "$dx$" may sound like a nice time (and paper) saving step. But it tends to lead to wrong answers.  
