Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$. 
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{(x+4)}{\sqrt{6}})+C$
However the result is $\frac{arctan(\frac{(x+4)}{\sqrt{6}})}{\sqrt{6}}+C$
Why is my result wrong? I can't see any mistake.

Your first step is wrong $$ \frac{1}{6+(x+4)^2} \color{red}{\neq} \frac{6}{1+\frac{(x+4)^2}{6}}$$
Also, your substitution is wrong. You have substituted $$ \mathrm dx =\frac{1}{\sqrt 6} \mathrm du$$ Which actually is $$\mathrm du =\frac{1}{\sqrt 6} \mathrm dx$$
The given answer is correct.
