Answer: the original integral is:
∫ (3x - 2) / (x - 3)(2x - 1) dx = (7/5)ln|x - 3| + (1/10)ln|2x - 1| + C
Step-by-step explanation:In order to solve for A and B, we need to expand the expression (x - 3)(2x - 1):
(x - 3)(2x - 1) = 2x^2 - x - 6x + 3
= 2x^2 - 7x + 3
Now, we equate the coefficients of like terms:
2x^2 - 7x + 3 = A(3x - 2)
Comparing the x^2 coefficients:
2 = 3A
A = 2/3
Comparing the x coefficients:
-7 = -2A
-7 = -2(2/3)
-7 = -4/3
-7 = B
Therefore, A = 2/3 and B = -7.
Now, let's evaluate the integral ∫ (3x - 2) / (x - 3)(2x - 1) dx:
We can start by performing a partial fractions decomposition:
(3x - 2) / (x - 3)(2x - 1) = A / (x - 3) + B / (2x - 1)
Multiplying both sides by (x - 3)(2x - 1), we get:
(3x - 2) = A(2x - 1) + B(x - 3)
Expanding and equating coefficients, we have:
3x - 2 = (2A + B)x - (A + 3B)
Equate the x coefficients:
3 = 2A + B
Equate the constant terms:
-2 = -A - 3B
Now we can solve these equations simultaneously:
From the first equation, we can solve for B:
B = 3 - 2A
Substituting into the second equation:
-2 = -A - 3(3 - 2A)
-2 = -A - 9 + 6A
-2 = 5A - 9
7 = 5A
A = 7/5
Substituting the value of A back into the equation for B:
B = 3 - 2(7/5)
B = 3 - 14/5
B = (15 - 14) / 5
B = 1/5
Therefore, A = 7/5 and B = 1/5.
Now we can rewrite the integral:
∫ (3x - 2) / (x - 3)(2x - 1) dx = ∫ (7/5)/(x - 3) dx + ∫ (1/5)/(2x - 1) dx
To integrate the first term, we use the natural logarithm function:
∫ (7/5)/(x - 3) dx = (7/5)ln|x - 3| + C
To integrate the second term, we use another substitution:
Let u = 2x - 1.
Then, du = 2dx, or dx = (1/2)du.
The integral becomes:
∫ (1/5)/(2x - 1) dx = ∫ (1/5)/u (1/2) du
= (1/10)ln|u| + C
= (1/10)ln|2x - 1| + C
Therefore, the original integral is:
∫ (3x - 2) / (x - 3)(2x - 1) dx = (7/5)ln|x - 3| + (1/10)ln|2x - 1| + C
i hope I'm right