<s>[INST] Where do I go wrong when solving this integral involving a quadratic? Simple question. I have an arithmetic mistake somewhere in here but can't find it.
$\int \frac{3x-1}{x^2+10x+28}dx$
$\frac{1}{3}\int \frac{3x-1}{\frac{1}{3}(x+5)^2+1}dx$
Set: 
$u=x+5$
$x=u-5$
$dx = du$
$\frac{1}{3}\int \frac{3u-16}{\frac{u^2}{3}+1}du$
$\frac{1}{3}\int \frac{3u}{\frac{u^2}{3}+1}du - \frac{1}{3}\int\frac{16}{\frac{u^2}{3}+1}du$
Set:
$u = \sqrt3 tan\theta$
$du = \sqrt3 sec^2\theta d\theta$
$\theta = arctan(\frac{u}{\sqrt3})$
$3\int tan\theta d\theta - \frac{1}{3}\int 16\sqrt3d\theta$
$3 ln(sec(arctan(\frac{u}{\sqrt3}))) - \frac{16}{\sqrt3}(arctan(\frac{u}{\sqrt3}))+c  $
$3 ln(\frac{\sqrt{u^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{u}{\sqrt3})+c$
$3 ln(\frac{\sqrt{(x+5)^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{x+5}{\sqrt3})+c$
$\frac{3}{2}ln(\frac{(x+5)^2}{3})-\frac{16}{\sqrt3}arctan(\frac{x+5}{\sqrt3})+c$
I think it's correct but there shouldn't be a $3$ in the denominator in the first term.  But I can't see where I went wrong.
 [/INST] Note that $\ln(g(x)/c)=ln(g(x))-ln(c)$ for any function $g(x)$. Since you're finding an antiderivative, any extra added/subtracted constants are irrelevant.
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