Evaluate the integral (3x^2 − 19x + 46) / (2x + 1)(x − 2)^2 dx

please help! it would be greatly appreciated :)

Since you can't factor the numerator, I'd go with partial fraction decomposition on this one.

A/(2x + 1) + B/(x - 2) + C/(x - 2)² = (3x^2 − 19x + 46) / (2x + 1)(x − 2)²
A(x - 2)² + B(2x + 1)(x - 2) + C(2x + 1) = 3x² - 19x + 46

Letting x = 2...
C*5 = 12 - 38 + 46
5C = 20
C = 4

Letting x = -1/2...
A(25/4) = 3(1/4) - 19(-1/2) + 46
25A = 3 + 38 + 46(4)
A = 9

Letting x = 0...
A(4) + B(1)(-2) + C(1) = 46
-2B = 46 - (4)(1) - (9)(4)
B = -3

So now you want to integrate 9/(2x + 1) - 3/(x - 2) + 4/(x - 2)², which if you can't do in your head, I would recommend u-substitution. The answer is (9/2)*ln|2x + 1| - 3ln|x - 2| - 4/(x - 2) + C

Check: http://www.wolframalpha.com/input/?i=int…

partial fractions.

(3x^2-19x+46)/(2x+1)(x-2)^2=A/(2x+1) + B/(x-2) + C/(x-2)^2

multiply both sides by (2x+1)(x-2)^2

3x^2-19x+46=A(x-2)(x-2)+B(2x+1)(x-2)+C…
plug in x=2
20=5C
C=4
evaluate coefficients.
x^2:
3=A+2B
x:
-19=-4A-3B+2C
x^0
46=4A-2B+C
C=4
-27=-4A-3B
42=4A-2B
add
15=-5B
B=-3
3=A+2B
3=A-6
A=9

A=9, B=-3, C=4
9/(2x+1) + -3/(x-2) + 4/(x-2)^2

integrate each term
(9/2)ln(2x+1)-3ln(x-2)-4/(x-2) + C.