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Math Help - Integration by parts confusion

  1. #1
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    Integration by parts confusion

    Hello,

    I have a pretty simple integration by parts question I'm supposed to solve, but for some reason I can't seem to get the right answer...Can someone point out where I've gone wrong?

    \int x\, ln x \, dx

    So, u = ln x \,

    dv= x \, dx \,

    du= \frac {1}{x} \,dx

    v = \frac {x^2}{2}

    Our tutor instructed us that if there is a natural log in the question, that that should be set as the variable to be deriviated, as we haven't been taught a technique for finding the integral of ln x.

    Anyway, I worked the question out as follows...

    = uv - \int v \, du

    = ln\,x (\frac {x^2}{2}) - \int \frac {x^2}{2} \, \frac {1}{x} \, dx

    = \frac {x^2 ln \,x}{2} - \frac {1}{2} \int x^2 \, \frac {1}{x} \,dx

    = \frac {x^2 ln \,x}{2} - \frac {1}{2} \cdot \frac {x^3}{3} \, ln\,x\,+\,C

    = \frac {x^2 ln \,x}{2} - \frac {x^3 \,ln\,x}{6}\,+\,C

    Apparently, I'm supposed to be getting this though...

    = \frac {x^2 ln \,x}{2} - \frac {x^2}{4} \,+\,C

    Where exactly am I going wrong?
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  2. #2
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    Re: Integration by parts confusion

    \int x^2\frac{1}{x}dx you integrated using a nonexisting rule \int f\cdot g=\int f \cdot \int g.
    You shoud be doing \int x^2\frac{1}{x}dx=\int x dx=\frac{x^2}{2}
    Thanks from astuart
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  3. #3
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    Re: Integration by parts confusion

    Quote Originally Posted by astuart View Post
    Hello,

    I have a pretty simple integration by parts question I'm supposed to solve, but for some reason I can't seem to get the right answer...Can someone point out where I've gone wrong?

    \int x\, ln x \, dx

    So, u = ln x \,

    dv= x \, dx \,

    du= \frac {1}{x} \,dx

    v = \frac {x^2}{2}

    Our tutor instructed us that if there is a natural log in the question, that that should be set as the variable to be deriviated, as we haven't been taught a technique for finding the integral of ln x.

    Anyway, I worked the question out as follows...

    = uv - \int v \, du

    = ln\,x (\frac {x^2}{2}) - \int \frac {x^2}{2} \, \frac {1}{x} \, dx

    = \frac {x^2 ln \,x}{2} - \frac {1}{2} \int x^2 \, \frac {1}{x} \,dx

    = \frac {x^2 ln \,x}{2} - \frac {1}{2} \cdot \frac {x^3}{3} \, ln\,x\,+\,C
    \int f(x)g(x)dx is NOT " \int f(x)dx\int g(x)dx"! Instead just note that \frac{x^2}{2}\frac{1}{x}= \frac{x}{2}.

    = \frac {x^2 ln \,x}{2} - \frac {x^3 \,ln\,x}{6}\,+\,C

    Apparently, I'm supposed to be getting this though...

    = \frac {x^2 ln \,x}{2} - \frac {x^2}{4} \,+\,C

    Where exactly am I going wrong?
    Thanks from astuart
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  4. #4
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    Re: Integration by parts confusion

    Sorry, but I don't quite understand why the integral sign is there then? Are you essentially saying that the integral sign should be ignored, and I should instead multiply \frac {x^2}{2} \cdot \frac {1}{x} resulting in \frac {x}{2}??

    I'm missing something really simple here, but it just hasn't clicked...
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  5. #5
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    Re: Integration by parts confusion

    Quote Originally Posted by astuart View Post
    Sorry, but I don't quite understand why the integral sign is there then? Are you essentially saying that the integral sign should be ignored, and I should instead multiply \frac {x^2}{2} \cdot \frac {1}{x} resulting in \frac {x}{2}??

    I'm missing something really simple here, but it just hasn't clicked...
    Nevermind, I see what you mean now....

    \frac {x^2\,lnx}{2} - \int \frac{x}{2} (from multiplying the two terms BEFORE taking the integral)

    = \frac {x^2\,lnx}{2} - \frac {1}{2} \int x

    = \frac {x^2\,lnx}{2} - \frac {1}{2} \cdot \frac{x^2}{2} \,+\,C

    = \frac {x^2\,lnx}{2} - \frac {x^2}{4}\,+\,C

    Thank you very much, both of you!!
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