# 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.

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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.
So it's not wrong, but I could have went one step further and said $c_1 = c + \frac{3}{2}ln(3)$? I don't know why I didn't see that. Thanks. –  Korgan Rivera Oct 18 '12 at 12:55