∫(x^4 /(x^3 −2x^2 −7x+4))dx


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Question Number 97591 by M±th+et+s last updated on 08/Jun/20

$\int \frac{x^{4}}{x^{3} - 2 x^{2} - 7 x + 4} d x$
	Answered by MJS last updated on 08/Jun/20

	$\int \frac{x^{4}}{x^{3} - 2 x^{2} - 7 x + 4} d x = \int (x + 2) d x + \int \frac{11 x^{2} + 10 x - 8}{x^{3} - 2 x^{2} - 7 x + 4} d x$ $x^{3} - 2 x^{2} - 7 x + 4 = (x - α) (x - α) (x - γ)$ $α = \frac{2}{3} - \frac{10}{3} \cos (\frac{π}{6} + \frac{1}{3} \arcsin \frac{17}{125})$ $β = \frac{2}{3} - \frac{10}{3} \sin (\frac{1}{3} \arcsin \frac{17}{125})$ $γ = \frac{2}{3} + \frac{10}{3} \sin (\frac{π}{3} + \frac{1}{3} \arcsin \frac{17}{125})$ $\int \frac{11 x^{2} + 10 x - 8}{x^{3} - 2 x^{2} - 7 x + 4} d x =$ $= \int (\frac{A}{x - α} + \frac{B}{x - β} + \frac{C}{x - γ}) d x =$ $= A \ln ∣ x - α ∣ + B \ln ∣ x - β ∣ + C \ln ∣ x - γ ∣$ $A = \frac{11 α^{2} + 10 α - 8}{(α - β) (α - γ)}$ $B = \frac{11 β^{2} + 10 β - 8}{(β - α) (β - γ)}$ $C = \frac{11 γ^{2} + 10 γ - 8}{(γ - α) (γ - β)}$ $no easier path$
	Commented by M±th+et+s last updated on 08/Jun/20

	Commented by bobhans last updated on 09/Jun/20

	$golden . . job$
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