Question

Please Help
Multiply (x – 4)(x2 + 5x + 6).

x3 + 7x2 – 20x – 24
x3 + x2 – 14x – 24
x3 + 9x2 – 26x – 24
x3 + 5x2 + 10x – 24

Answer 1

∫ [(x - 4) /(x² - 2x - 3x + 6)] dx =

∫ {(x - 4) /[x(x - 2) - 3(x - 2)]} dx =

∫ {(x - 4) /[(x - 2)(x - 3)]} dx =

then decompose the integrand into partial fractions:

(x - 4) /[(x - 2)(x - 3)] = A/(x - 2) + B/(x - 3)

(x - 4) /[(x - 2)(x - 3)] = [A(x - 3) + B(x - 2)] /[(x - 2)(x - 3)]

x - 4 = A(x - 3) + B(x - 2)

x - 4 = Ax - 3A + Bx - 2B

x - 4 = (A + B)x + (- 3A - 2B)

A + B = 1
- 3A - 2B = - 4

A = 1 - B
3A + 2B = 4

A = 1 - B
3(1 - B) + 2B = 4

A = 1 - B
3 - 3B + 2B = 4

A = 1 - B
- B = 4 - 3

A = 1 - B
- B = 1

A = 1 - (- 1) = 1 + 1 = 2
B = - 1

hence:

(x - 4) /[(x - 2)(x - 3)] = A/(x - 2) + B/(x - 3) = 2/(x - 2) - 1/(x - 3)

thus the integral becomes:

∫ [(x - 4) /(x² - 5x + 6)] dx = ∫ {[2/(x - 2)] - [1/(x - 3)]} dx =

break it up pulling constants out:

2 ∫ [1 /(x - 2)] dx - ∫ [1 /(x - 3)] dx =

2 ln |x - 2| - ln |x - 3| + C (antiderivative)

having the antiderivative, plug in the bounds:

1
∫ [(x - 4) /(x² - 5x + 6)] dx = (2 ln |1 - 2| - ln |1 - 3|) - (2 ln |0 - 2| - ln |0 - 3|) =
0

2 ln |- 1| - ln |- 2| - 2 ln |- 2| + ln |- 3| =

2 ln (1) - ln (2) - 2 ln (2) + ln (3) =

2(0) - 3 ln (2) + ln (3) =

(recalling logarithm properties)

ln (3) - ln (2³) =

ln (3) - ln (8) =

ln (3/8) ≈ - 0.980829

Answer 2

I don't exactly understand that