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Question Number 21662 by x² - y²@gmail.com last updated on 30/Sep/17

Answered by Joel577 last updated on 30/Sep/17

I = ∫ ((8x^5  − 18x^3 )/(2x^3  − 3x^2 )) dx = ∫ ((2x^3 (4x^2  − 9))/(x^2 (2x − 3))) dx     = ∫ ((2x(2x + 3)(2x − 3))/(2x − 3)) dx     = ∫ 4x^2  + 6x dx     = (4/3)x^3  + 3x^2  + C

$${I}\:=\:\int\:\frac{\mathrm{8}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{3}} }{\mathrm{2}{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}^{\mathrm{2}} }\:{dx}\:=\:\int\:\frac{\mathrm{2}{x}^{\mathrm{3}} \left(\mathrm{4}{x}^{\mathrm{2}} \:−\:\mathrm{9}\right)}{{x}^{\mathrm{2}} \left(\mathrm{2}{x}\:−\:\mathrm{3}\right)}\:{dx} \\ $$$$\:\:\:=\:\int\:\frac{\mathrm{2}{x}\left(\mathrm{2}{x}\:+\:\mathrm{3}\right)\left(\mathrm{2}{x}\:−\:\mathrm{3}\right)}{\mathrm{2}{x}\:−\:\mathrm{3}}\:{dx} \\ $$$$\:\:\:=\:\int\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:{dx} \\ $$$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{C} \\ $$

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