How about the substitution y = 1+kx => dy = k dx <=> 1/k dy = dx?
Hi there...
Okay, so... I'm trying to find out why...
Integral of 1 / (1 + k * x) dx
[1] results in (from Maple)... ln(1 + k * x)/k?
I've tried to use substitution (with help from Maple), which leads to...
u = 1/k + x
du = 1
What I suppose would lead to the following...
Integral of 1 / (k * u) du
[2] = LN| (1/k) + x | * (1/k)
Now... [1] and [2] are not similar... Can anyone explain what I'm not doing right?
Thanks
Simon DK
Hello, sh01by!
I didn't understand your steps at all . . .
. . then I figured out what Maple did!
Results ifrom Maple: . . . . right!
I've tried to use substitution (with help from Maple), which leads to...
? ? . What is that? . (Oh, I see!)
Let
Substitute: .
Back-substitute: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's what Maple did . . . Ha!
Let
Substitute: .
Back-substitute: .
Now watch this . . .
We have: .
. .
Therefore, we have: . . . . . ta-DAA!
The good news:. Your answer is correct!