∫ ((x^4 −5x^3 +6x^2 −18)/(x^3 −3x^2 )) dx ?


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Question Number 121696 by bemath last updated on 11/Nov/20

$\int \frac{x^{4} - 5 x^{3} + 6 x^{2} - 18}{x^{3} - 3 x^{2}} d x ?$
	Answered by liberty last updated on 11/Nov/20
	$J=∫ [(x−2)−((18)/(x^2 (x−3))) ]dx = J=(1/2)(x−2)^2 −18∫ (1/(x^2 (x−3))) dx Partial fraction (1/(x^2 (x−3))) = (A/x) + (B/x^2 ) + (C/(x−3)) C = [ (1/x^2 ) ]_(x=3) = (1/9) B = [ (1/(x−3)) ]_(x=0) = −(1/3) ⇔1 = Ax(x−3) −(1/3)(x−3)+(1/9)x^2 x=1⇒1 = −2A +(2/3)+(1/9) ⇒A=−(1/9) J = (((x−2)^2 )/2)−18∫ (−(1/(9x))+(1/(9(x−3)))−(1/(3x^2 ))) dx J = (((x−2)^2 )/2)+2ln ∣x∣ −2ln ∣x−3∣ −(6/x) + c . ▲$
	$J = \int [(x - 2) - \frac{18}{x^{2} (x - 3)}] dx =$ $J = \frac{1}{2} {(x - 2)}^{2} - 18 \int \frac{1}{x^{2} (x - 3)} dx$ $Partial fraction$ $\frac{1}{x^{2} (x - 3)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x - 3}$ $C = {[\frac{1}{x^{2}}]}_{x = 3} = \frac{1}{9}$ $B = {[\frac{1}{x - 3}]}_{x = 0} = - \frac{1}{3}$ $\Leftrightarrow 1 = Ax (x - 3) - \frac{1}{3} (x - 3) + \frac{1}{9} x^{2}$ $x = 1 \Rightarrow 1 = - 2 A + \frac{2}{3} + \frac{1}{9} \Rightarrow A = - \frac{1}{9}$ $J = \frac{{(x - 2)}^{2}}{2} - 18 \int (- \frac{1}{9 x} + \frac{1}{9 (x - 3)} - \frac{1}{{3 x}^{2}}) dx$ $J = \frac{{(x - 2)}^{2}}{2} + 2 \ln ∣ x ∣ - 2 \ln ∣ x - 3 ∣ - \frac{6}{x} + c . ▴$
	Commented by bemath last updated on 11/Nov/20

	$t y p o i t s h o u l d b e$ $1 = A x (x - 3) - \frac{1}{3} (x - 3) + \frac{1}{9} x^{2}$ $x = 1 \Rightarrow 1 = - 2 A + \frac{2}{3} + \frac{1}{9}$ $\Rightarrow 2 A = - \frac{2}{9}; A = - \frac{1}{9}$ $J = \frac{{(x - 2)}^{2}}{2} - 18 \int (- \frac{1}{9 x} - \frac{1}{3 x^{2}} + \frac{1}{9 (x - 3)}) d x$ $J = \frac{{(x - 2)}^{2}}{2} + 2 \ln ∣ x ∣ - \frac{6}{x} - 2 \ln ∣ x - 3 ∣ + c$ $J = \frac{{(x - 2)}^{2}}{2} + \ln ∣ \frac{x}{x - 3} ∣^{2} - \frac{6}{x} + c$
	Answered by MJS_new last updated on 11/Nov/20

	$\int \frac{x^{4} - 5 x^{3} + 6 x^{2} - 18}{x^{3} - 3 x^{2}} d x =$ $= \int x - 2 + \frac{2}{x} - \frac{2}{x - 3} + \frac{6}{x^{2}} d x =$ $= \frac{x^{2}}{2} - 2 x + 2 \ln ∣ x ∣ - 2 \ln ∣ x - 3 ∣ - \frac{6}{x} =$ $= \frac{x^{3} - 4 x^{2} - 12}{2 x} + 2 \ln ∣ \frac{x}{x - 3} ∣ + C$
	Commented by liberty last updated on 11/Nov/20

	$Ostrogradski$
	Commented by MJS_new last updated on 11/Nov/20

	$no . just decomposition .$
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