#x^3 + x^2 = x^2(x + 1)#
So,
#(Ax + B)/x^2 + C/(x+ 1) = (x^2 + 33)/((x^2)(x + 1))#
#(Ax + B)(x + 1) + Cx^2 = x^2 + 33#
#Ax^2 + Bx + Ax +B + Cx^2 = x^2 + 33#
#(A + C)x^2 + (A + B)x + B = x^2 + 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!
