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 – 8×2 = 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) —— […]

Integration 1

Differentiate y = 5xe2x+1 with respect to x and hence, find ∫x(e2x+1) dx. Solution: 1/2(x)(e2x+1) – 1/4(e2x+1) + C dy/dx = 5x(e2x+1)(2) + 5(e2x+1)           =10x(e2x+1) + 5(e2x+1) 5x(e2x+1) = ∫10x(e2x+1) + 5(e2x+1) dx 5x(e2x+1) = 10∫x(e2x+1) dx + 5∫(e2x+1) dx 10∫x(e2x+1) dx + 5/2(e2x+1) = 5x(e2x+1) 10∫x(e2x+1) dx = 5x(e2x+1) -5/2(e2x+1) ∫x(e2x+1) dx […]