$\int (\frac{2x-1}{x^2-6x+13})dx$
ให้ $u=x^2-6x+13 , du=(2x-6)dx , dx=\frac{du}{2x-6}$
$\int (\frac{2x-1}{x^2-6x+13})dx=\int\frac{2x-1}{u}\cdot \frac{du}{2x-6}$
$=\int\frac{(2x-6)+5}{u}\cdot \frac{du}{2x-6}$
$=\int \frac{1}{u}du+\int\frac{5}{(2x-6)u}du$
$=\int \frac{1}{u}du+5\int\frac{1}{x^2-6x+13}dx$
$=\int \frac{1}{u}du+5\int\frac{1}{(x-3)^2+4}dx$
$=\int \frac{1}{u}du+\frac{5}{4}\int\frac{1}{\frac{(x-3)^2+4}{4}}dx$
$=\int \frac{1}{u}du+\frac{5}{2}\int\frac{1}{(\frac{x}{2}-\frac{3}{2})^2+1}d(\frac{x}{2}-\frac{3}{2})$
$=\ln(x^2-6x+13)+\frac{5}{2}\arctan[\frac{x}{2}-\frac{3}{2}]+c$
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