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Question Number 86039 by ar247 last updated on 26/Mar/20

∫((2x^5 −x^3 −1)/(x^3 −4x))dx

$$\int\frac{\mathrm{2}{x}^{\mathrm{5}} −{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$

Commented by abdomathmax last updated on 27/Mar/20

A =∫ ((2x^2 (x^3 −4x)+7x^3 −1)/(x^3 −4x))dx =∫ 2x^2 dx +∫ ((7x^3 −1)/(x^3 −4x))dx =(2/3)x^3 +∫ ((7(x^3 −4x)+28x−1)/(x^3 −4x))dx =(2/3)x^3 +7x +∫ ((28x−1)/(x^3 −4x))dx let devompose F(x) =((28x−1)/(x(x^2 −4))) =((28x−1)/(x(x−2)(x+2))) =(a/x) +(b/(x−2)) +(c/(x+2)) a =(1/4) b=((55)/(2.4)) =((55)/8) c =((−57)/((−2)(−4))) =((57)/8) ⇒ ∫ F(x)dx =(1/4)ln∣x∣+((55)/8)ln∣x−2∣ +((57)/8)ln∣x+2∣ +C ⇒ A =(2/3)x^3 +7x +(1/4)ln∣x∣ +((55)/8)ln∣x−2∣ +((57)/8)ln∣x+2∣ +C

$${A}\:=\int\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{4}{x}\right)+\mathrm{7}{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\int\:\mathrm{2}{x}^{\mathrm{2}} \:{dx}\:+\int\:\:\frac{\mathrm{7}{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:+\int\:\:\frac{\mathrm{7}\left({x}^{\mathrm{3}} −\mathrm{4}{x}\right)+\mathrm{28}{x}−\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:+\mathrm{7}{x}\:\:+\int\:\:\frac{\mathrm{28}{x}−\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{4}{x}}{dx}\:\:{let}\:{devompose} \\ $$$${F}\left({x}\right)\:=\frac{\mathrm{28}{x}−\mathrm{1}}{{x}\left({x}^{\mathrm{2}} −\mathrm{4}\right)}\:=\frac{\mathrm{28}{x}−\mathrm{1}}{{x}\left({x}−\mathrm{2}\right)\left({x}+\mathrm{2}\right)} \\ $$$$=\frac{{a}}{{x}}\:+\frac{{b}}{{x}−\mathrm{2}}\:+\frac{{c}}{{x}+\mathrm{2}} \\ $$$${a}\:=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${b}=\frac{\mathrm{55}}{\mathrm{2}.\mathrm{4}}\:=\frac{\mathrm{55}}{\mathrm{8}} \\ $$$${c}\:=\frac{−\mathrm{57}}{\left(−\mathrm{2}\right)\left(−\mathrm{4}\right)}\:=\frac{\mathrm{57}}{\mathrm{8}}\:\Rightarrow \\ $$$$\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{x}\mid+\frac{\mathrm{55}}{\mathrm{8}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{57}}{\mathrm{8}}{ln}\mid{x}+\mathrm{2}\mid\:+{C}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} \:+\mathrm{7}{x}\:\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\mid{x}\mid\:+\frac{\mathrm{55}}{\mathrm{8}}{ln}\mid{x}−\mathrm{2}\mid\:+\frac{\mathrm{57}}{\mathrm{8}}{ln}\mid{x}+\mathrm{2}\mid\:+{C} \\ $$

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