Question #804a1

1 Answer
Mar 30, 2017

int x^3/(x^2-x-2)dx = x^2/2+x +1/3ln abs(x+1) +8/3 ln abs (x-2) +C

Explanation:

Consider the integrand function:

x^3/(x^2-x-2)

Factorize the denominator:

x^2-x-2 = (x+1)(x-2)

Now add ans subtract 1 to the numerator:

x^3/(x^2-x-2) = (x^3+1-1)/((x+1)(x-2)) = (x^3+1)/((x+1)(x-2)) -1/((x+1)(x-2))

Use the factorization: (x^3+1) = (x+1)(x^2-x+1) so that we can simplify:

(1) x^3/(x^2-x-2) = ( (x+1)(x^2-x+1) )/((x+1)(x-2)) -1/((x+1)(x-2)) = (x^2-x+1) /(x-2) -1/((x+1)(x-2))

Now note that:

x^2-x+1 = x^2-x-2 +3 = (x-2)(x+1) +3

so :

(x^2-x+1) /(x-2) = ((x-2)(x+1) +3)/(x-2) = x+1 +3/(x-2)

and that using partial fractions:

1/((x+1)(x-2)) = A/(x+1) +B/(x-2)

1/((x+1)(x-2)) = (Ax-2A+Bx+B)/((x+1)(x-2))

{(A+B = 0),(-2A+B = 1):}

{(A = -1/3),(B = 1/3):}

Substitute this in (1) to have:

x^3/(x^2-x-2) =x+1 +3/(x-2) +1/(3(x+1))-1/(3(x-2))

x^3/(x^2-x-2) =x+1 +1/(3(x+1))+8/(3(x-2))

We can now integrate:

int x^3/(x^2-x-2)dx = x^2/2+x +1/3ln abs(x+1) +8/3 ln abs (x-2) +C
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