Find the integral of 2/((2-x)(x+2)^2)dx.
I tried to use partial fractions to do it, but it didn't work.
Of course it works. Why don't you show your work and the reason why you rejected it?
2/((2-x)(x+2)^2) = (1/8) (1/(x+2) + 4/(x+2)^2 - 1/(x-2))
So, the integral is just
(1/8) (ln(x+2) - 4/(x+2) - ln(x-2))
= (1/8) (ln((x+2)/(x-2)) - 4/(x+2)) + C
To find the integral of 2/((2-x)(x+2)^2), we can use a technique called partial fractions.
First, we need to factor the denominator (2-x)(x+2)^2, which gives us:
(2-x)(x+2)^2 = (2-x)(x+2)(x+2)
Next, we express the given fraction as a sum of two or more simpler fractions with unknown numerators:
2/((2-x)(x+2)^2) = A/(2-x) + B/(x+2) + C/(x+2)^2
To find the unknowns A, B and C, we can multiply both sides of the equation by the denominator (2-x)(x+2)^2 to clear the fractions.
2 = A(x+2)^2 + B(2-x)(x+2) + C(2-x)
We can simplify this equation by expanding, collecting like terms, and equating coefficients.
Now, let's solve for A, B, and C:
For x = 2, the equation becomes:
2 = A(4)^2 + B(0) + C(0) = 16A
So, A = 2/16 = 1/8.
For x = -2, the equation becomes:
2 = A(0) + B(4)(0) + C(4) = 4C
So, C = 2/4 = 1/2.
Now, we can solve for B by substituting any other value that is convenient. Let's use x = 0:
2 = A(2)^2 + B(2)(2) + C(2) = 4A + 4B + 2C
2 = 4(1/8) + 4B + 2(1/2)
2 = 1/2 + 4B + 1
2 - 1/2 - 1 = 4B
3/2 = 4B
B = 3/8.
Now that we have found the values of A, B, and C, we can replace them in the equation:
2/((2-x)(x+2)^2) = 1/8/(2-x) + 3/8/(x+2) + 1/2/(x+2)^2.
Now, we can integrate each term separately:
∫(1/8/(2-x)) dx = (-1/8)ln|2-x| + C1,
∫(3/8/(x+2)) dx = (3/8)ln|x+2| + C2,
∫(1/2/(x+2)^2) dx = (-1/2)/(x+2) + C3.
Finally, the integral of 2/((2-x)(x+2)^2)dx is:
(-1/8)ln|2-x| + (3/8)ln|x+2| - (1/2)/(x+2) + C,
where C is the constant of integration.