How do you integrate (x^3+2)/(x^2-x)x3+2x2−x using partial fractions?

1 Answer
Narad T.
Oct 18, 2016

=x^2/2+x-2lnx+3ln(x-1)+C=x22+x−2lnx+3ln(x−1)+C

Explanation:

First make a long division to determine
(x^3+2)/(x^2-x)=x+1+(x+2)/(x^2-x)x3+2x2−x=x+1+x+2x2−x
To simplify the fraction we use partial fraction
(x+2)/(x^2-x)=(x+2)/(x(x-1))=A/x+B/(x-1)=(A(x-1)+Bx)/(x(x-1))x+2x2−x=x+2x(x−1)=Ax+Bx−1=A(x−1)+Bxx(x−1)
So we determine A and B
x+2=A(x-1)+Bxx+2=A(x−1)+Bx
Cmparing the coefficients, we find
1=A+B1=A+B and 2=-A2=−A and we conclude B=3B=3
int((x^3+2)dx)/(x^2-x)=int(x+1)dx-int2dx/x+int3dx/(x-1)∫(x3+2)dxx2−x=∫(x+1)dx−∫2dxx+∫3dxx−1
=x^2/2+x-2lnx+3ln(x-1)+C=x22+x−2lnx+3ln(x−1)+C

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