how do you integrate (3x+1)/(x-1)x^2 dx?

Answer 1

Is the x^2 also in the denominator?
In other words, do you want the integral of
(3x+1)/[(x-1)x^2] dx ?

I will assume that is what you mean.

Have you heard of the method of partial fractions? That is what you need to use here.

Your integrand can be rewritten as the sum of three terms:
A/x + B/x^2 + C/(x-1)

Figure out what A, B and C are, and integrate and sum the three simpler terms separately.

[Ax(x-1) + B(x-1) + C x^2]/[x^2(x-1)]
= (3x+1)/[(x-1)x^2]
A + C = 0
-B = 1 ; B = -1
B -A = 3 ; A = -4
C = 4

So your integral is the integral of
-4/x -1/x^2 + 4/(x-1)

(if I did the math right). In any case, that is the method to use.

Answer 2

yes, I got this answer, as well
THANKS