Question Number 113867 by Ar Brandon last updated on 15/Sep/20

∫_3 ^6 ((x+1)/(x^3 +x^2 −6x))dx

Answered by abdomsup last updated on 15/Sep/20

I =∫_3 ^6  ((x+1)/(x^3  +x^2 −6x))dx let decompise  F(x)=((x+1)/(x^3  +x^2 −6x)) ⇒  F(x) =((x+1)/(x(x^2  +x−6)))  x^2  +x−6=0→Δ=1+24=25  x_1 =((−1+5)/2)=2 and x_2 =((−1−5)/2)=−3  ⇒F(x)=((x+1)/(x(x−2)(x+3)))  =(a/x) +(b/(x−2)) +(c/(x+3))  a=−(1/6) , b=(3/(2(6))) =(1/4)  c=((−2)/((−3)(−5))) =(2/(15)) ⇒  F(x)=−(1/(6x)) +(1/(4(x−2))) +(2/(15(x+3)))  ⇒I =∫_3 ^6 (−(1/(6x)) +(1/(4(x−2)))+(2/(15(x+3))))dx  =−(1/6)[ln∣x∣]_3 ^6  +(1/4)[ln∣x−2∣]_3 ^6   +(2/(15))[ln∣x+3∣]_3 ^6   =−(1/6)ln(2)+(1/4)(2ln2)  +(2/(15)){2ln(3)−ln(6)}  =((1/2)−(1/6))ln(2)+(2/(15)){ln(3)−ln(2)}  =((1/3)−(2/(15)))ln(2)+(2/(15))ln(3)  =(1/5)ln(2)+(2/(15))ln(3)