Hmm, so if I have |(1+y)/(1-y)|, and I consider case 1, where -1<y<1 I get that |1+y| > 0 and |1-y|> 0, so the whole expression is positive, so then A must be > 0. So then I guess I have (1+y) = (1-y)Ae^2x, so y + yAe^2x = Ae^2x - 1, so y(1 +Ae^2x) = Ae^2x -1 so y = (Ae^2x -1)/(Ae^2x + 1)...
Hmm, so if I have |(1+y)/(1-y)|, and I consider case 1, where -1<y<1 I get that |1+y| > 0 and |1-y|> 0, so the whole expression is positive, so then A must be > 0. So then I guess I have (1+y) = (1-y)Ae^2x, so y + yAe^2x = Ae^2x - 1, so y(1 +Ae^2x) = Ae^2x -1 so y = (Ae^2x -1)/(Ae^2x + 1).
When...
I have this equation: dy/dx = 1-y^2, so then dy/(1-y^2) = dx, so ∫dy/(1-y^2) = dx --->
∫(A/(1+y) + B/(1-y))dy = x + C. I rewrite it again: (A - Ay + B + By)/(1+y)(1-y) = 1/(1+y)(1-y) so I get A+B = 1, and B-A = 0, so B = A, and therefore 2A = 1. so A & B = 1/2.
So 1/2∫(dy/(1+y) +...
Alright, I am here again with another question...
When I have a rational function, let's say (x+4)/(x-2)(x-3) I rewrite it like A/(x-2) + B(x-3) and then solve it for A & B. But when we have for e.g (x^2 + 3x + 2)/(x(x^2 +1 )) the book tells me to rewrite it like:
A/x + (Bx + C)/(x^2 + 1)...
Hey!
In my calculus book they claim that a second degree polynomial always can be rewritten as x^2 - a^2 or as x^2 + a^2, if you use an appropriate change of variable. I was thinking about how this works.
Let's say we have a second degree polynomial (on the general form?) ax^2 +bx + c = 0...
Hey!
I have this integral: ∫((1/2)/(2x-1))dx.
The first time, I did like this: ∫((1/2)/(2x-1))dx = (1/2)∫(1/(2x-1))dx. If I set u = 2x-1, then du = 2dx, so I can rewrite (1/2)∫(1/(2x-1))dx as (1/2)*(1/2)∫(1/u)du = 1/4∫(1/u)du = 1/4ln|u| = 1/4ln|2x-1|.
But when I do like this (I cannot...