Integrate $\int\frac{2x+6}{x^2+6x+18}\mathrm dx$

Question

My problem is: $$\int\frac{2x+6}{x^2+6x+18}\mathrm dx$$ This is part of a larger integral so I didn't see the $y=x^2+6x+18$ substitution straight away. My concern is what was wrong with my working in the first place: I substituted $3u=x+3$ to get $$\int\frac{6u}{9u^2+9}\cdot 3\mathrm du$$ $$=\int\frac{2u}{u^2+1}\mathrm du$$ $$=\int\frac{\mathrm dv}{v}$$ where $v=u^2+1$ $$=\ln(u^2+1)+C\\=\ln\left(\frac{x^2}{9}+\frac{2x}{3}+2\right)+C$$ However, if I go back to my original observation that I can substitute $y=x^2+6x+18$, I get $$\int\frac{\mathrm dy}{y}\\=\ln(x^2+6x+18)+C$$

What did I do wrong the first time?

Your integrand is of the form $\frac{f'(x)}{f(x)}$, so you do not really need all the calculations to arrive at the result $\ln(f(x))+C$. — Joonas Ilmavirta, Aug 10 '14 at 10:42
http://math.stackexchange.com/questions/892986/how-to-compute-int-fracxx2-4x82-dx — lab bhattacharjee, Aug 10 '14 at 14:21

Peter Woolfitt · Accepted Answer · 2014-08-10T10:45:13.840

3

The answers are the same. Note that

$$\ln\left(\frac{x^2}{9}+\frac{2x}{3}+2\right)+C=\ln\left(\frac{x^2}{9}+\frac{2x}{3}+2\right)+\ln(9)+C'=\ln(x^2+6x+18)+C'$$

where $C'$ is an arbitrary constant offset from $C$ by $\ln(9)$.

edited Aug 10 '14 at 10:45

answered Aug 10 '14 at 10:40

Peter Woolfitt

21,476

Integrate $\int\frac{2x+6}{x^2+6x+18}\mathrm dx$

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