Solve the integral (x+3)/(x-1)^3
So far, I'm stuck. This is what I have thus far:
x+3 = A(x-1)^3/(x-1) + B(x-1)^3/(x-1)^2 + C(x-1)^3/(x-1)^3
x+3 = A(x^2-2x+1) + B(x-1) + C
x+3 = Ax^2 - 2Ax +Bx + A - B + C
x+3 =Ax^2 + 1(B-2A)x + (A - B + C)
let u = x-1
(x+3)/(x-1)^3 = (u + 4) / u^3 = u^(-2) + 4 u^(-3)
Okay, I understand how to solve now, but I'm curious on where did that four come from?
Thank you.
distributive property: (u+4)*v = uv + 4v
duh
x - 1 = u ... x + 3 = u + 4
To solve the integral of (x+3)/(x-1)^3, you can use the method of partial fractions.
First, let's factor the denominator (x-1)^3:
(x-1)^3 = (x-1)(x-1)(x-1)
To decompose the fraction into partial fractions, we assume that the numerator can be expressed as the sum of fractions with denominators (x-1), (x-1)^2, and (x-1)^3:
(x+3) / (x-1)^3 = A/(x-1) + B/(x-1)^2 + C/(x-1)^3
Now, let's find the values of A, B, and C.
Multiplying both sides by (x-1)^3, we get:
(x+3) = A(x-1)^2 + B(x-1) + C
Expanding the right side:
x+3 = A(x^2-2x+1) + B(x-1) + C
Rearranging terms:
x+3 = Ax^2 - 2Ax + A + Bx - B + C
Now, let's group the x-terms and constant terms separately:
x+3 = Ax^2 + (B-2A)x + (A - B + C)
For both sides of the equation to be equal, the coefficients of x and the constants must be equal.
x-term:
1 = B - 2A
Constant term:
3 = A - B + C
We now have a system of equations with two unknowns (A and B).
From the x-term equation, we can solve for B:
B = 1 + 2A
Substituting this into the constant term equation:
3 = A - (1 + 2A) + C
3 = A - 1 - 2A + C
3 = -A - 2 + C
5 = -A + C
Now, we have two equations and two unknowns:
1 = B - 2A
5 = -A + C
Solving this system of equations, we find the values of A, B, and C.
First, let's solve for A:
From the first equation:
1 = 1 + 2A - 2A
1 = 1
This means that A can take any value.
Next, let's solve for B:
From the first equation:
B = 1 + 2A
Finally, let's solve for C:
By substituting the value of A and B into the second equation:
5 = -A + C
5 = -A + C
This means that C can also take any value.
Since A, B, and C can take any value, we can write the partial fraction decomposition as:
(x+3)/(x-1)^3 = A/(x-1) + B/(x-1)^2 + C/(x-1)^3
The integral of each term in the partial fraction decomposition can be found easily:
∫A/(x-1) dx = Aln|x-1| + constant
∫B/(x-1)^2 dx = -B/(x-1) + constant
∫C/(x-1)^3 dx = -C/2(x-1)^2 + constant
Therefore, the final solution to the integral ∫(x+3)/(x-1)^3 dx is:
Aln|x-1| - B/(x-1) - C/2(x-1)^2 + constant, where A, B, C, and constant are constants that depend on the initial conditions.