Question

Hello, I would really appreciate some help with the following question. Please show steps. Thankyou

∫ (-5z)/(z²+5) dz

Answer 1

ok
Alright, first consider the following system from the coefficients:

8x + 5y + 3 = 0
5x + 2y + 4 = 0

Solving gives x = -14/9 and y = 17/9. Thus the substitution we have to make is

u = x - (-14/9) = x + 14/9 ====> x = u - 14/9
v = y - 17/9 =============> y = v + 17/9

dy/dx = v' / u' = dv/du

dv/du = [8(u - 14/9) + 5(v + 17/9) + 3] / [5(u - 14/9) + 2(v + 17/9) + 4]
dv/du = [8u - 112/9 + 5v + 85/9 + 3] / [5u - 70/9 + 2v + 34/9 + 4]
dv/du = [8u - 112/9 + 5v + 85/9 + 3] / [5u - 70/9 + 2v + 34/9 + 4]
dv/du = [8u + 5v] / [5u + 2v]
v' = [8u + 5v] / [5u + 2v]

q = v/u ==> u = v/q
v = qu
v' = q + uq' = q + (v/q)q'

q + (v/q)q' = [8(v/q) + 5v] / [5(v/q) + 2v]
q + (v/q)q' = [5q + 8] / [2q + 5]
q^2 + vq' = [(q)(5q + 8)] / [2q + 5]
vq' = [(q)(5q + 8)] / [2q + 5] - q^2
vq' = (8q - 2q^3) / (2q + 5)
v(dq/dv) = (8q - 2q^3) / (2q + 5)
v dq = (8q - 2q^3) / (2q + 5) dv
(2q + 5) / (8q - 2q^3) dq = (1/v) dv

Finally we can integrate this. The left one is partial fractions all the way:

∫ (2q + 5) / (8q - 2q^3) dq = ∫ (1/v) dv
(1/16) [-9 ln(2 - q) + 10 ln(q) - ln(q + 2)] = ln(v) + C
(1/16) [-9 ln(2 - v/u) + 10 ln(v/u) - ln(v/u + 2)] = ln(v) + C

v/u = (y - 17/9) / (x + 14/9)

(1/16) [-9 ln((18 x - 9 y + 45)/(9 x + 14)) + 10 ln((y - 17/9) / (x + 14/9)) - ln((18 x + 9 y + 11)/(9 x + 14))] = ln(y - 17/9 ) + C

Answer 2

I'm not getting it. Could you plz put our reply more in the form of my question?