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Simple question. I have an arithmetic mistake somewhere in here but can't find it.

$\int \frac{3x-1}{x^2+10x+28}dx$

$\frac{1}{3}\int \frac{3x-1}{\frac{1}{3}(x+5)^2+1}dx$




$dx = du$

$\frac{1}{3}\int \frac{3u-16}{\frac{u^2}{3}+1}du$

$\frac{1}{3}\int \frac{3u}{\frac{u^2}{3}+1}du - \frac{1}{3}\int\frac{16}{\frac{u^2}{3}+1}du$


$u = \sqrt3 tan\theta$

$du = \sqrt3 sec^2\theta d\theta$

$\theta = arctan(\frac{u}{\sqrt3})$

$3\int tan\theta d\theta - \frac{1}{3}\int 16\sqrt3d\theta$

$3 ln(sec(arctan(\frac{u}{\sqrt3}))) - \frac{16}{\sqrt3}(arctan(\frac{u}{\sqrt3}))+c $

$3 ln(\frac{\sqrt{u^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{u}{\sqrt3})+c$

$3 ln(\frac{\sqrt{(x+5)^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{x+5}{\sqrt3})+c$


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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up vote 3 down vote accepted

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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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
I also have on occasion been thrown off by log terms in antiderivatives not matching those of an "answer at the back of the book". It's always a case of log(c*h(x))... – coffeemath Oct 18 '12 at 13:23

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