<s>[INST] 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.
 [/INST] 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:
$$\frac{1}{\sqrt6} arctan(\frac{x+4}{\sqrt6}) + C$$
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