how do you integrate (3x+1)/(x-1)x^2 dx?
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.
yes, I got this answer, as well
THANKS
To integrate the given expression, we can use the method of partial fractions. Here is how you can do it step by step:
1. First, factorize the denominator: x^3 - x^2
x^3 - x^2 = x^2(x - 1)
2. Write the given expression as a sum of two fractions:
(3x + 1)/(x - 1)x^2 = A/x^2 + B/(x - 1)
3. Find the values of A and B. To do this, we need to cross multiply and solve for A and B.
(3x + 1) = A(x - 1) + Bx^2
Expand the right side: 3x + 1 = Ax - A + Bx^2
Rearrange the terms: Bx^2 + (A - 3)x + (-A - 1) = 0
Now, equate the coefficients of x^2, x, and the constant term on both sides.
Coefficient of x^2: B = 0
Coefficient of x: A - 3 = 3
Constant term: -A - 1 = 0
From the second equation, we get A = 6, and from the third equation, we get A = -1.
So, A = -1 and B = 0.
4. Now, rewrite the integral using the partial fractions:
∫ [(3x + 1)/((x - 1)x^2)] dx = ∫[-1/x^2 + 0/(x - 1)] dx
5. Integrate each fraction separately using the power rule:
∫ -1/x^2 dx = -(-1)/x = 1/x
∫ 0/(x - 1) dx = 0
6. Combine the results:
∫ [(3x + 1)/((x - 1)x^2)] dx = 1/x + C
Therefore, the integral of (3x + 1)/((x - 1)x^2) with respect to x is 1/x + C, where C is the constant of integration.