Why can't I use the form $\frac1{x^2+1}$ from the derivative of $\arctan(x)$ to convert the integral form in this situation? In the question that solves the integral $\displaystyle\int\frac1{6x^2 + 36x + 78} \,\mathrm{d}x$, I first tried to solve it by changing the denominator in a form of $\dfrac1{x^2 + 1}$ to apply $\;\arctan(x)$.
$\dfrac1{6\!\cdot\!\left(x^2 + 6x + 13\right)}=\dfrac16\!\cdot\!\dfrac1{(x + 3)^2 + 2^2}$
Now, in order to make it in a form of $\;\dfrac1{x^2 + 1}\;,\;$ I divide everything by $\,2^2$:
$=\dfrac16\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}\!\cdot\!\dfrac1{2^2}
=\dfrac1{24}\!\cdot\!\dfrac1{\left(\frac{x+3}2\right)^2+1}$
Then assume that $\;u=\dfrac{x+3}2\;,\;$ I thought I can apply arctan to get rid of the integral form:
$\dfrac1{24}\!\cdot\!\arctan\left(\dfrac{x+3}2\right)+c\;.$
But the correct answer is $\;\dfrac1{12}\!\cdot\!\arctan\left(\dfrac{x+3}2\right) + c\;,\;$ not $\;\dfrac1{24}\,.$
The answer also explained to use $\;\dfrac1{x^2 + k^2} = \left[\dfrac1k\!\cdot\!\arctan\left(\dfrac xk\right)\right]’$ (derivative), but I wonder why I cannot use the form of $\;\dfrac1{x^2 + 1}\;,\;$ which is the only equation I have known.

Basically, you messed up in your substitution:
$$\int \frac{1}{6x^2+36x+78}dx = \frac{1}{6} \int \frac{1}{(x+3)^2+2^2}dx =  \frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx$$
Let $u=\frac{x+3}{2}$, Then $du = \frac{dx}{2}$ So:
$$\frac{1}{24} \int \frac{1}{(\frac{x+3}{2})^2+1}dx = \frac{1}{24} \int \frac{1}{(u)^2+1} \cdot 2 du = \frac{1}{12} \int \frac{1}{(u)^2+1} du $$
$$= \frac{1}{12}\arctan(\frac{x+3}{2})$$
It seems you instead used $du=dx$
Note that you can use whatever form you'd like to evaluate your integral but what the question tells you to use requires less working so may aswell.
