∫ ((4x−1)/(2x^2 −3x+2)) dx ?


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Question Number 123793 by liberty last updated on 28/Nov/20

$\int \frac{4 x - 1}{2 x^{2} - 3 x + 2} d x ?$
	Answered by Dwaipayan Shikari last updated on 28/Nov/20

	$\int \frac{4 x - 3}{2 x^{2} - 3 x + 2} + 2 \int \frac{1}{2 x^{2} - 3 x + 2} d x$ $= l o g (2 x^{2} - 3 x + 2) + \int \frac{1}{{(x - \frac{3}{4})}^{2} + {(\sqrt{1 - \frac{9}{16}})}^{2}}$ $= l o g (2 x^{2} - 3 x + 2) + \frac{4}{\sqrt{7}} {t a n}^{- 1} \frac{4 x - 3}{\sqrt{7}} + C$
	Answered by liberty last updated on 28/Nov/20
	$partial fraction ((4x−1)/(2x^2 −3x+2)) = P (((d/dx)(2x^2 −3x+2))/(2x^2 −3x+2))+(Q/(2x^2 −3x+2)) 4x−1 = P(4x−3)+Q ; { ((P=1)),((−1=Q−3P ;Q=2)) :} I(x)=∫((4x−3)/(2x^2 −3x+2)) dx + ∫ (2/(2(x^2 −((3x)/2)+1)))dx I(x)= ℓn ∣2x^2 −3x+2∣ + ∫ (dx/((x−(3/4))^2 +(((√7)/4))^2 )) I(x)= ℓn ∣2x^2 −3x+2∣+(4/( (√7))) arctan (((x−(3/4))/((√7)/4))) +c I(x)=ln ∣2x^2 −3x+2∣ +(4/( (√7)))arctan (((4x−3)/( (√7))))+c$
	$p a r t i a l f r a c t i o n \frac{4 x - 1}{2 x^{2} - 3 x + 2} = P \frac{\frac{d}{d x} (2 x^{2} - 3 x + 2)}{2 x^{2} - 3 x + 2} + \frac{Q}{2 x^{2} - 3 x + 2}$ $4 x - 1 = P (4 x - 3) + Q; {\begin{cases} P = 1 \\ - 1 = Q - 3 P; Q = 2 \end{cases}$ $I (x) = \int \frac{4 x - 3}{2 x^{2} - 3 x + 2} d x + \int \frac{2}{2 (x^{2} - \frac{3 x}{2} + 1)} d x$ $I (x) = ℓ n ∣ 2 x^{2} - 3 x + 2 ∣ + \int \frac{d x}{{(x - \frac{3}{4})}^{2} + {(\frac{\sqrt{7}}{4})}^{2}}$ $I (x) = ℓ n ∣ 2 x^{2} - 3 x + 2 ∣ + \frac{4}{\sqrt{7}} \arctan (\frac{x - \frac{3}{4}}{\frac{\sqrt{7}}{4}}) + c$ $I (x) = \ln ∣ 2 x^{2} - 3 x + 2 ∣ + \frac{4}{\sqrt{7}} \arctan (\frac{4 x - 3}{\sqrt{7}}) + c$
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