Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 10

int(x^3-x+3)/(x^2+x-2)dx
The given integrand is a improper rational function, as the degree of the numerator is more than the degree of the denominator. To apply the method of partial fractions,first we have to do a division with remainder.
(x^3-x+3)/(x^2+x-2)=(x-1)+(2x+1)/(x^2+x-2)
Since the polynomials do not completely divide, we have to continue partial fractions on the remainder.
We need to factor the denominator,
(2x+1)/(x^2+x-2)=(2x+1)/(x^2-x+2x-2)
=(2x+1)/(x(x-1)+2(x-1))
=(2x+1)/((x-1)(x+2))
Now let's create the partial fraction template,
(2x+1)/((x-1)(x+2))=A/(x-1)+B/(x+2)
Multiply the above equation by denominator,
=>2x+1=A(x+2)+B(x-1)
2x+1=Ax+2A+Bx-B
2x+1=(A+B)x+2A-B
Equating the coefficients of the like terms,
A+B=2 ------------------------(1)
2A-B=1 -------------------------(2)
Now we have to solve the above two linear equations to get A and B,
Add the equations 1 and 2,
A+2A=2+1
3A=3
A=1
Plug in the value of A in equation 1,
1+B=2
B=2-1=1
Now plug in the values of A and B in the partial fraction template,
(2x+1)/((x-1)(x+2))=1/(x-1)+1/(x+2)
Now we can evaluate the integral as,
int(x^3-x+3)/(x^2+x-2)dx=int((x-1)+1/(x-1)+1/(x+2))dx
Apply the sum rule,
=intxdx-int1dx+int1/(x-1)dx+int1/(x+2)dx
For the first and second integral apply the power rule and for the third and fourth integral use the common integral:int1/xdx=ln|x|
=x^2/2-x+ln|x-1|+ln|x+2|
Add a constant C to the solution,
=x^2/2-x+ln|x-1|+ln|x+2|+C