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3.2(B) · Q80

Q.Evaluate: ∫14x2−20x+17 dx\int \frac{1}{4x^2-20x+17}\,dx

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Factor: 4x2−20x+17=4(x2−5x)+17=4(x−52)2−25+17=4(x−52)2−8=4[(x−52)2−2]4x^2-20x+17=4\left(x^2-5x\right)+17=4\left(x-\dfrac52\right)^2-25+17=4\left(x-\dfrac52\right)^2-8=4\left[\left(x-\dfrac52\right)^2-2\right].

So ∫dx4x2−20x+17=14∫duu2−(2)2\displaystyle\int\dfrac{dx}{4x^2-20x+17}=\dfrac14\int\dfrac{du}{u^2-(\sqrt2)^2} with u=x−52u=x-\dfrac52: …

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