` int[3x+5]/[x^3-x^2-x+1].dx`

1.	` int[3x+5]/[x^3-x^2-x+1].dx`
Answer» `(x^(3)-x^(2)-x-1)=x^(2)(x-1)-(x-1)-=(x-1)(x^(2)-1)=(x-1)^(2)(x+1).` `Let (3x+5)/((x^(3)-x^(2)-x-1))=(3x+5)/((x-1)^(2)(x+1))=(A)/((x-1))+(B)/((x-1)^(2))+( C)/((x+1))` `implies (3x+5)-=A(x-1)(x+1)+B(x+1)+C(x-1)^(2).` Putting `x=1` on both sides of (i) we get `B=4.` Putting `x=-1` on both sides of `(i)`, we get `C=(1)/(2).` Comparing the coefficient of `x^(2)` on both sides of (i) , we get `A+C=0implies A=-C=(-1)/(2)` `therefore ((3x+5))/((x^(3)+x^(2)+x+1))=(-1)/(2(x-1))+(4)/((x-1)^(2))+(1)/(2(x+1))` `implies ((3x+5))/((x^(3)-x^(2)-x+1))dx=-(1)/(2)int(dx)/((x-1))+4int(dx)/((x-1))^(2)+(1)/(2)int(dx)/((x+1))=-(1)/(2)log\|x-1\|-(4)/((x-1))+(1)/(2)log\|x+1\|+C.`

Discussion

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