<s>[INST] 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?
 [/INST] As you noted, if $u = 4x$, then $du = 4dx$, yielding
$$\begin{split}
\int \frac{12dx}{16x^2+1} 
 &= 3 \int \frac{4dx}{(4x)^2+1} \\
 &= 3 \int \frac{u'dx}{(u)^2+1} \\
 &= 3 \int \frac{du}{u^2+1} \\
 &= 3 \arctan(u) + C \\
 &= 3 \arctan(4x)+C
\end{split}
$$
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