Solving integration by the use of partial fractions

Solution:

(a) A=2, B=-5, C=0

(b) 2x/(x2+3)

(c) ln(2x-1) – (5/2)ln(x2+3) + C

(a)6 + 5x – 8x2 = A(x2 + 3) + (Bx + c)(2x – 1)

Sub x = ½

6 + 5/2 – 8(1/2)2 = A(1/4 + 3) + 0

13/2 = (13/4)A —— A=2

When x=0, 6 = 2(3) + c(-1) —— C=0

Compare coefficients of x2

-8 = A + 2B

-10 – 2B —— B = -5

 

(b)d/dx[ln(x2+3)] = 2x/(x2+3)

=2x/(x2+3)

 

(c)∫ 2/(2x-1) + ∫(-5x)/(x2+3)

=ln(2x-1) – (5/2)ln(x2+3) + C

 

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