x^3 + x^2 = x^2(x + 1)x3+x2=x2(x+1)
So,
(Ax + B)/x^2 + C/(x+ 1) = (x^2 + 33)/((x^2)(x + 1))Ax+Bx2+Cx+1=x2+33(x2)(x+1)
(Ax + B)(x + 1) + Cx^2 = x^2 + 33(Ax+B)(x+1)+Cx2=x2+33
Ax^2 + Bx + Ax +B + Cx^2 = x^2 + 33Ax2+Bx+Ax+B+Cx2=x2+33
(A + C)x^2 + (A + B)x + B = x^2 + 33(A+C)x2+(A+B)x+B=x2+33
So,
{(A + C = 1), (A + B = 0), (B = 33):}
Solving, we get that A = -33, B = 33, C = 34.
Hence, the partial fraction decomposition is as follows:
(-33x + 33)/x^2 + 34/(x + 1)
We can integrate the second term as 34ln|x + 1| + C. However, we can decompose the first term further.
int(33- 33x)/x^2dx = int(33(1 - x))/x^2dx = 33int(1- x)/x^2dx
A/x + B/x^2 = (1 - x)/x^2
Ax + B = 1 - x
Here, we have that A = -1 and B = 1.
Hence,
=>33int(1/x^2 - 1/x)dx
=>33int(x^(-2) - 1/x)dx
=>33(-1/x - ln|x|) + C
Putting this and the part that we integrated above together gives
=>-33/x - 33ln|x| + 34ln|x + 1| + C
Hopefully this helps!