1. ## Simple integration

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

2. How about the substitution y = 1+kx => dy = k dx <=> 1/k dy = dx?

3. Originally Posted by TD!
How about the substitution y = 1+kx => dy = k dx <=> 1/k dy = dx?
It also depends whether $k=0$ and $k\not = 0$. Two seperate integrations.

4. Hello, sh01by!

I didn't understand your steps at all . . .
. . then I figured out what Maple did!

$\int \frac{1}{1 + kx}\,dx$

Results ifrom Maple: . $\frac{1}{k}\ln(1 + k x) + C$ . . . right!

I've tried to use substitution (with help from Maple), which leads to...

$u \:= \:\frac{1}{k} + x$ ? ? .
What is that? . (Oh, I see!)

Let $u \:=\:1 + kx\quad\Rightarrow\quad du = k\,dx\quad\Rightarrow\quad dx = \frac{du}{k}$

Substitute: . $\int \frac{1}{u}\left(\frac{du}{k}\right)\;=\;\frac{1}{ k}\int\frac{du}{u}\;=\;\frac{1}{k}\ln|u| + C$

Back-substitute: . $\boxed{\frac{1}{k}\ln|1 + kx| + C}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's what Maple did . . . Ha!

$\int\frac{1}{1+kx}\,dx\;=\;\int\frac{1}{k\left(\fr ac{1}{k} + C\right)}\,dx\;=\;\frac{1}{k}\int\frac{1}{\frac{1} {k} + x}\,dx$

Let $u \:=\:\frac{1}{k} + x\quad\Rightarrow\quad du \,=\,dx$

Substitute: . $\frac{1}{k}\int\frac{1}{u}\,du \;=\;\frac{1}{k}\ln|u| + C$

Back-substitute: . $\frac{1}{k}\ln\left|\frac{1}{k} + x\right| + C$

Now watch this . . .

We have: . $\frac{1}{k}\ln\left|\frac{1}{k} + x\right| + C \;\;=\;\;\frac{1}{k}\ln\left|\frac{1 + kx}{k}\right| + C$

. . $=\;\frac{1}{k}\bigg[\ln|1+kx| - \ln|k|\bigg] + C\;\;=\;\;\frac{1}{k}\ln|1 + kx| - \underbrace{\frac{1}{k}\ln|k| + C}_{\text{This is a constant!}}$

Therefore, we have: . $\boxed{\frac{1}{k}\ln\left|1 + kx\right| + C}$ . . . . ta-DAA!

5. ## Thanks!

Thanks, I was getting a little tired of Maple's tutor (which otherwise is very helpful to people like me)... : o )

Have a nice evening...

Simon DK

6. Originally Posted by ThePerfectHacker
It also depends whether $k=0$ and $k\not = 0$. Two seperate integrations.
It was $k\not = 0$... I'll write all details next time... It was part of a larger integration process...

Thanks though...