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

∫((2x^5 −x^3 −1)/(x^3 −4x))dx

∫2x5−x3−1x3−4xdx

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=∫2x2(x3−4x)+7x3−1x3−4xdx=∫2x2dx+∫7x3−1x3−4xdx=23x3+∫7(x3−4x)+28x−1x3−4xdx=23x3+7x+∫28x−1x3−4xdxletdevomposeF(x)=28x−1x(x2−4)=28x−1x(x−2)(x+2)=ax+bx−2+cx+2a=14b=552.4=558c=−57(−2)(−4)=578⇒∫F(x)dx=14ln∣x∣+558ln∣x−2∣+578ln∣x+2∣+C⇒A=23x3+7x+14ln∣x∣+558ln∣x−2∣+578ln∣x+2∣+C

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