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Thread: Integration by sustitution

  1. #1
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    Integration by substitution

    I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

    the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

    I can't seem to bash it out though.

    u=1/x

    du/dx=-1/x^2

    when I try to substitute it all goes to pot.

    I need the method before moving onto the real work!

    Thanks

    TF
    Last edited by Taylor Fylde; May 14th 2008 at 12:35 AM.
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  2. #2
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    Quote Originally Posted by Taylor Fylde View Post
    I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

    the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

    I can't seem to bash it out though.

    u=1/x

    du/dx=-1/x^2

    when I try to substitute it all goes to pot.

    I need the method before moving onto the real work!

    Thanks

    TF
    It's easier to do it using partial fractions, but if you insist:

    $\displaystyle du = -\frac{1}{x^2}~dx$

    $\displaystyle dx = -x^2~du$

    $\displaystyle dx = -\frac{1}{u^2}~du$

    So
    $\displaystyle \int \frac{1}{x(1 - ax)}~dx = -\int \frac{u}{u^2 \left ( 1 - \frac{a}{u} \right ) }~du$

    $\displaystyle = -\int \frac{1}{u - a}~du$

    $\displaystyle = -ln \left | u - a \right | + C$

    $\displaystyle = -ln \left | \frac{1}{x} - a \right | + C$

    -Dan
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by Taylor Fylde View Post
    I want to integrate 1/x(1-ax) wrt to x. It can be done using the substitution u=1/x

    the answer is y=C-ln|(1/x)-a| where C is an arbitrary constant.

    I can't seem to bash it out though.

    u=1/x

    du/dx=-1/x^2

    when I try to substitute it all goes to pot.

    I need the method before moving onto the real work!

    Thanks

    TF
    Ok for the substitution... Let's study it :

    $\displaystyle \boxed{\frac{1}{x(1-ax)}}$

    As you said : $\displaystyle \frac{du}{dx}=-\frac{1}{x^2}$

    But we want it with respect to u.

    $\displaystyle \frac{du}{dx}=-\frac{1}{x^2}=-\left(\frac 1x\right)^2=-u^2$

    $\displaystyle \implies dx=-\frac{du}{u^2}$

    Hence, the substitution yields :

    $\displaystyle \begin{aligned} \int \frac{1}{x(1-ax)} dx &= \int \frac 1x \cdot \frac{1}{1-ax} dx \\
    &= \int u \cdot \frac{1}{1-\frac au} \cdot \left(-\frac{1}{u^2}\right) du \\
    &= -\int \frac 1u \cdot \frac{1}{1-\frac au} du \\
    &= -\int \frac{1}{u(1-\frac au)} du \end{aligned}$
    $\displaystyle \begin{aligned} &= -\int \frac{1}{u-a} du \\
    &= -\ln \left|u-a\right| + c \\
    &= -\ln \left|\frac 1x-a\right| + c
    \end{aligned}$
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  4. #4
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    Many thanks (both of you).

    I'm getting it, but I seem to have to substitute a bit at a time. Each time I write it out, I substitute in a different order. Time for bed - let it ferment over night.

    TF
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