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3.2(B) · Q84

Q.Evaluate: ∫18−3x+2x2 dx\int \frac{1}{\sqrt{8-3x+2x^2}}\,dx

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Write 8−3x+2x2=2x2−3x+8=2(x2−32x+4)8-3x+2x^2=2x^2-3x+8=2\left(x^2-\dfrac32x+4\right), so integral =12∫dxx2−32x+4=\dfrac{1}{\sqrt2}\int\dfrac{dx}{\sqrt{x^2-\frac32x+4}}.

Complete the square: x2−32x+4=(x−34)2+5516x^2-\dfrac32x+4=\left(x-\dfrac34\right)^2+\dfrac{55}{16}. …

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