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.
 A: You can write it shortly as $$\int  \frac { 1 }{ 6+\left( x+4 \right) ^{ 2 } } dx=\frac { 1 }{ 6 } \int  \frac { \sqrt { 6 } d\left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  }{ 1+{ \left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  }^{ 2 } } =\frac { \sqrt { 6 }  }{ 6 } \arctan { \left( \frac { x+4 }{ \sqrt { 6 }  }  \right)  } +C$$
A: 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.
A: Start U-substitution:
$u$ = $x$+4 and $dx$= 1$dx$ $$\\$$
Making the integral:
$$\int \frac{1}{6+(u)^2} du$$ $$\\$$
Now do another U-substitution:
$u$ = $\sqrt6v$  and $du$= $\sqrt6 dv$ $$\\$$
Making the integral: 
$$\int \frac{1}{\sqrt6+(v^2+1)} dv$$ $$\\$$
Take out the constant:
$$\frac{1}{\sqrt6}\int \frac{1}{(v^2+1)} dv$$
The integrand is a common integral. 
The integral of $\frac{1}{(v^2+1)} dv$ is $arctan(v)$.
So now you have:
$\frac{1}{\sqrt6} arctan(v)$
Don't forget to substitute back in $v$ and $u$.
$v$= $\frac{u}{\sqrt6}$ and $u$= $x$+4. $$\\$$
Your final answer is:
$$\frac{1}{\sqrt6} arctan(\frac{x+4}{\sqrt6}) + C$$
A: Anyway everyone should know that
$$\int\frac{\mathrm dx}{x^2+a^2}=\frac1a\arctan\frac xa.$$
