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Thread: Integration Help

  1. #1
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    Integration Help

    hey there..
    i have an assignment to do and dont really know how to do this question...

    $\displaystyle \int x^n ln(ax) dx$ (n cant equal -1)

    do i do it by substitution? parts???
    where do i start..???
    if someone could help would be great. thanks
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  2. #2
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    Just directly apply integration by parts starting with:
    $\displaystyle \begin{array}{lll} u = \ln (ax) & \quad & dv = x^{n} dx \\ du = \text{???} & \quad & v = \text{???}\end{array}$

    Then use the fact that: $\displaystyle \int u dv = uv - \int v du$
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  3. #3
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    what would du be in this case??? not very great with logs...

    would it be 1/(ax) ???
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    You must've learned about the chain rule right?

    $\displaystyle \left[f(g(x))\right]' = f'(g(x)) \cdot g'(x)$

    Imagine $\displaystyle f(x) = \ln (x)$ and $\displaystyle g(x) = ax$
    $\displaystyle \left[\ln (ax)\right]' = \frac{1}{ax} \cdot (ax)'$

    etc. etc.
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  5. #5
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    yes. i have

    i got this far and now stuck again. dont know waht to do..

    x^n ln(ax) dx

    Let u = ln(ax)
    du = 1/ax . a = 1/x
    dv = x^n dx
    v = x^(n+1) / (n+1)


    udv = uv - vdu

    ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1)) / (n+1) . 1/x

    how do i proceed now??
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    Quote Originally Posted by b00yeah05 View Post
    yes. i have

    i got this far and now stuck again. dont know waht to do..

    x^n ln(ax) dx

    Let u = ln(ax)
    du = 1/ax . a = 1/x
    dv = x^n dx
    v = x^(n+1) / (n+1)


    udv = uv - vdu

    ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1)) / (n+1) . 1/x

    how do i proceed now??
    The integration by parts formula is $\displaystyle \int u \, dv = uv - \int v \, du$.

    After making the appropriate substitutions, the integral on the right hand side should be blue sky all the way ......
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  7. #7
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    sorry i knew the forumla just didnt know how to type it using the latex thing. i have multiplied them together and dont know how to proceed...

    on the right hand side i get...

    x^(n+1) / x(n+1)

    i know that the x's cancel out but then how do i integrate the restt?? and how do i simplify the left hand side?
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  8. #8
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    Ok so making your substitutions into the formula:
    $\displaystyle \int x^{n} \ln (ax) dx = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \int \left(\frac{x^{n+1}}{n+1} \cdot \frac{1}{x} \right) dx$

    $\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1}\int \frac{x^{n+1}}{x} dx$ (Factored out the constant $\displaystyle \frac{1}{n+1}$ )

    Remember that: $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$

    etc. etc.
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  9. #9
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    sorry would it be possible for you to do the rest of the working and give an answer pl.ease so i can understand?? i have never come accross a question like this..i have really tried to do it but i cant get around it.
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  10. #10
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    Quote Originally Posted by b00yeah05 View Post
    sorry would it be possible for you to do the rest of the working and give an answer pl.ease so i can understand?? i have never come accross a question like this..i have really tried to do it but i cant get around it.
    Are you saying you can't find $\displaystyle \int \frac{x^{n+1}}{x} dx$?
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  11. #11
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    yes...that and i dont know how to simplify the left hand side...
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  12. #12
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    Do you realise that $\displaystyle \int \frac{x^{n+1}}{x} dx = \int x^n\, dx
    $ ?

    The LHS is already simplified
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  13. #13
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    yes i know that. although i dont know WHY...so i was just looking for the working for that...i cant seem to get around it. therefore because of the factor that was taken out before....would it just be (x^(n+1)) / (n+1)^2 ????

    therefore the full answer would be....

    ln(ax) . (x^(n+1)) / (n+1) - (x^(n+1))/ (n+1)^2

    ?? am i correcT?
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  14. #14
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    yes i know that. although i dont know WHY
    As I told you: $\displaystyle \frac{a^{m}}{a^{n}} = a^{m-n}$
    Applying this to your case: $\displaystyle \frac{x^{n+1}}{x^{1}} = x^{n+1-1} = x^{n}$

    --------------

    $\displaystyle \int x^{n} \ln (ax) = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1}\int \frac{x^{n+1}}{x} dx$
    $\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \int x^{n} dx$
    $\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{1}{n+1} \cdot \frac{x^{n+1}}{n+1} + c$
    $\displaystyle = \ln (ax) \left(\frac{x^{n+1}}{n+1}\right) - \frac{x^{n+1}}{(n+1)^{2}} + c$

    which agrees with your answer.
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  15. #15
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    ahhhhh!!!!! i was thinking the x's cancel thats why it was making no sense to me....thanks for the help though...
    great help
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