Calculate y when dy/dx = 1/(bx+5)^3

Answer 1

dy = (bx+5)^-3 dx

let z = bx+5
then dz = b dx and dx = (1/b) dz

dy = (1/b) z^-3 dz

y = (1/b)(-1/2)z^-2 + c

y = -1/(2b z^2) + c
but z^2 = b^2x^2 + 10 bx + 25
so
2 b z^2 = 2b^3x^2 + 20 b^2 x + 50 b
so
y = -1/(2b^3x^2 + 20 b^2 x + 50 b) + c

Answer 2

or, given the original form of the function,

y = -1/(2(bx+5)^2)

Answer 3

I have a extra b in there.

Answer 4

actually, I dropped a b!

y = -1/(2b(bx+5)^2)

Answer 5

Two b or not two b