By using partial fractions, integrate the function: f(x) = (4-2x)/(2x+1)(x+1)(x+3)

(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1) + B/(x+1) + C/(x+3) 4-2x = A(x+1)(x+3) + B(2x+1)(x+3) + C(2x+1)(x+1) let x = -1: 4-2(-1) = B(2(-1)+1)((-1)+3) 6 = B(-1)(2) B = -3 let x = -3: 4-2(-3)= C(2(-3)+1)((-3)+1) 10 = C(-5)(-2) C = 1 let x = -1/2: 4-2(-1/2) = A(-1/2 + 1)(-1/2 + 3) 5 = A(1/2)(5/2) A = 4 f(x) = 4/(2x+1) - 3/(x+1) + 1/(x+3) int(f(x)) = int(4/(2x+1)) dx - int(3/(x+1)) dx + int(1/(x+3)) dx = 2int(2/(2x+1))dx -3int(1/(x+1))dx + int(1/(x+3))dx = 2ln|2x+1| - 3ln|x+1| +ln|x+3| + c