Step 1: Match the first integral.
Complete the square: $x^2+4x+\alpha=(x+2)^2+(\alpha-4)$. The arctan answer $\dfrac{1}{2\sqrt2}\tan^{-1}\!\dfrac{x+2}{2\sqrt2}$ comes from $\int\dfrac{dx}{(x+2)^2+a^2}=\dfrac1a\tan^{-1}\dfrac{x+2}{a}$.
Step 2: Find $a$ and then $\alpha$.
Comparing, $a=2\sqrt2$ so $a^2=8$. Then $\alpha-4=8$, giving $\alpha=12$.
Step 3: Set up the second integral.
We now need $\displaystyle\int\frac{dx}{x^2+4x-12}$ (using $\alpha=12$ with a minus sign).
Step 4: Complete the square again.
\[ x^2+4x-12=(x+2)^2-16=(x+2)^2-4^2. \]
Step 5: Use the log formula.
Since $\displaystyle\int\frac{du}{u^2-a^2}=\frac{1}{2a}\log\left|\frac{u-a}{u+a}\right|$, with $u=x+2$ and $a=4$:
\[ \frac{1}{8}\log\left|\frac{(x+2)-4}{(x+2)+4}\right|. \]
Step 6: Simplify.
\[ \frac{1}{8}\log\left|\frac{x-2}{x+6}\right|+k. \]
\[ \boxed{\dfrac{1}{8}\log\left|\dfrac{x-2}{x+6}\right|+k} \]