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[Solved] Integral question

 
Integral
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June 15, 2021 2:50 pm
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(@armor)
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Find the value of   \int_{1}^{2}\frac{x^4-2x^2+3x-2}{x^2-6x+9}dx.

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June 15, 2021 2:51 pm
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(@armor)
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First we note that \small x^2-6x+9=(x-3)^2,  and so we write 

\small x^4-2x^2+3x-2=x^3(x-3)+3x^2(x-3)+7x(x-3)+24(x-3)+70

\small =(x-3)[x^2(x-2)+6x(x-3)+25(x-3)+99]+70

the integral therefore becomes

\small \int_{1}^{2}\left \{ x^2+6x+25+\frac{99}{x-3}+\frac{70}{(x-3)^2} \right \}dx

\small =[\frac{1}{3}x^3+3x^2+25x+99ln|x-3|-\frac{70}{(x-3)}]_{1}^{2}

\small =\frac{8}{3}+12+50+70+99ln1-(\frac{1}{3}+3+25+99ln2+353)=\frac{214}{3}-99ln2.

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