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.

You can write it shortly as $$\int  \frac { 1 }{ 6+\left( x+4 \right) ^{ 2 } } dx=\frac { 1 }{ 6 } \int  \frac { \sqrt { 6 } d\left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  }{ 1+{ \left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  }^{ 2 } } =\frac { \sqrt { 6 }  }{ 6 } \arctan { \left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  } +C$$
