Exercise:
Compute, $\int_{1}^{3}{\frac{\ln x}{{{x}^{2}}-3x+3}dx}$
Solution: Let $I=\int_{1}^{3}{\frac{\ln x}{{{x}^{2}}-3x+3}dx}$ and Put $x=\frac{3}{y}\Leftrightarrow dx=\frac{-3}{{{y}^{2}}}dy\Leftrightarrow y=\frac{3}{x}$
So $\int_{1}^{3}{\frac{\ln x}{{{x}^{2}}-3x+3}dx}=\int_{3}^{1}{\frac{\ln \left( \frac{3}{y} \right)}{{{\left( \frac{3}{y} \right)}^{2}}-3\left( \frac{3}{y} \right)+3}\times \frac{-3}{{{y}^{2}}}dy}$
$=3\int_{1}^{3}{\frac{\ln 3-\ln y}{9-9y+3{{y}^{2}}}dy}=\int_{1}^{3}{\frac{\ln 3}{{{y}^{2}}-3y+3}}\,dy-\int_{1}^{3}{\frac{\ln y}{{{y}^{2}}-3y+3}dy}$
$\Rightarrow 2\int_{1}^{3}{\frac{\ln x}{{{x}^{2}}-3x+3}dx}=\int_{1}^{3}{\frac{\ln 3}{{{y}^{2}}-3y+3}dy}=\ln 3\int_{1}^{3}{\frac{dy}{{{\left( y-\frac{3}{2} \right)}^{2}}+\frac{3}{4}}}$
Let $w=y-\frac{3}{2}\Leftrightarrow dw=dy$
Observe that , $\int{\frac{dy}{{{\left( y-\frac{3}{2} \right)}^{2}}+\frac{3}{4}}=\int{\frac{dw}{{{w}^{2}}+\frac{3}{4}}}=\int{\frac{1}{{{\left( \frac{\sqrt{3}}{2}u \right)}^{2}}+\frac{3}{4}}\times \frac{\sqrt{3}}{2}du}}$
$=\int{\frac{\sqrt{3}}{\frac{3}{2}\left( {{u}^{2}}+1 \right)}du}=\frac{2\sqrt{3}}{3}\arctan u+c=\frac{2\sqrt{3}}{3}\arctan \left( \frac{2}{\sqrt{3}}w \right)+c$
Thus $\int{\frac{dy}{{{y}^{2}}-3y+3}=\frac{2\sqrt{3}}{3}\arctan \left( \frac{2\sqrt{3}}{3}\left( y-\frac{2}{3} \right) \right)+c}$
Hence $\int_{1}^{3}{\frac{dy}{{{y}^{2}}-3y+3}}=\underset{y\to {{3}^{-}}}{\mathop{\lim }}\,\frac{2\sqrt{3}}{3}\arctan \left( \frac{2\sqrt{3}}{3}\left( y-\frac{3}{2} \right) \right)-\underset{y\to {{1}^{+}}}{\mathop{\lim }}\,\frac{2\sqrt{3}}{3}\arctan \left( \frac{2\sqrt{3}}{3}\left( y-\frac{3}{2} \right) \right)$
0 $=\frac{2\sqrt{3}}{3}\left( \arctan \left( \sqrt{3} \right)-\arctan \left( -\frac{\sqrt{3}}{3} \right) \right)=\frac{2\sqrt{3}}{3}\left( \frac{\pi }{3}+\frac{\pi }{6} \right)=\frac{\pi \sqrt{3}}{3}=\frac{\pi }{\sqrt{3}}$
Thus $\int_{1}^{3}{\frac{\ln x}{{{x}^{2}}-3x+3}dx}=\frac{\ln 3}{2}\times \frac{\pi }{\sqrt{3}}=\frac{\pi \ln 3}{2\sqrt{3}}$
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