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Math Help - Simple integration

  1. #1
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    Post 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
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  2. #2
    TD!
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    How about the substitution y = 1+kx => dy = k dx <=> 1/k dy = dx?
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  3. #3
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    Quote Originally Posted by TD! View Post
    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.
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  4. #4
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    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!


    The good news:. Your answer is correct!

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  5. #5
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    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
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  6. #6
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    Smile

    Quote Originally Posted by ThePerfectHacker View Post
    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...
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