Results 1 to 4 of 4

Math Help - Integration by sustitution

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    6

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,885
    Thanks
    325
    Awards
    1
    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:

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

    dx = -x^2~du

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

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

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

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

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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 :

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

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

    But we want it with respect to u.

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

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

    Hence, the substitution yields :

    \begin{aligned} \int \frac{1}{x(1-ax)} dx &= \int \frac 1x \cdot \frac{1}{1-ax} dx \\<br />
&= \int u \cdot \frac{1}{1-\frac au} \cdot \left(-\frac{1}{u^2}\right) du \\<br />
&= -\int \frac 1u \cdot \frac{1}{1-\frac au} du \\<br />
&= -\int \frac{1}{u(1-\frac au)} du \end{aligned}
    \begin{aligned} &= -\int \frac{1}{u-a} du \\<br />
&= -\ln \left|u-a\right| + c \\<br />
&= -\ln \left|\frac 1x-a\right| + c<br />
\end{aligned}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    6
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: May 20th 2011, 07:04 PM
  2. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  3. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. sustitution intregration again
    Posted in the Calculus Forum
    Replies: 8
    Last Post: June 8th 2007, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum