How do you integrate (x^2+x)/((x+2)(x-1)^2)x2+x(x+2)(x−1)2 using partial fractions?
1 Answer
int (x^2+x)/((x+2)(x-1)^2) dx =2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C∫x2+x(x+2)(x−1)2dx=29ln|x+2|+79ln|x−1|−23(x−1)+C
Explanation:
Since the denominator has already been factored for us, we can tell that we're looking for a partial fraction decomposition of the form:
(x^2+x)/((x+2)(x-1)^2)x2+x(x+2)(x−1)2
=A/(x+2)+B/(x-1)+C/(x-1)^2=Ax+2+Bx−1+C(x−1)2
=(A(x-1)^2+B(x+2)(x-1)+C(x+2))/((x+2)(x-1)^2)=A(x−1)2+B(x+2)(x−1)+C(x+2)(x+2)(x−1)2
=((A+B)x^2+(-2A+B+C)x+(A-2B+2C))/((x+2)(x-1)^2)=(A+B)x2+(−2A+B+C)x+(A−2B+2C)(x+2)(x−1)2
Equating coefficients we get the following system of linear equations:
{ (A+B=1), (-2A+B+C=1), (A-2B+2C=0) :}
Adding all three equations together, we find:
3C = 2
So
Subtracting the second equation from the first, we have:
3A-C = 0
Hence
Then from the first equation we find
So:
(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)
Hence:
int (x^2+x)/((x+2)(x-1)^2) dx
=int 2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2) dx
=2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C
