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Math Help - integration by part

  1. #1
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    integration by part

    I'm having some trouble with this

    \int{\ln{(2x+1)}dx}

    I set u = ln(2x+1) and dv = dx so that du=\frac{2}{2x+1} and v = x.

    so after integrating by part once I get:

    x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}

    then I tried to integrate by part again with u=x and dv=(2x+1)^{-1}

    my final answer ends up as x\ln{(2x+1)} but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by stevedave View Post
    I'm having some trouble with this

    \int{\ln{(2x+1)}dx}

    I set u = ln(2x+1) and dv = dx so that du=\frac{2}{2x+1} and v = x.

    so after integrating by part once I get:

    x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}

    then I tried to integrate by part again with u=x and dv=(2x+1)^{-1}

    my final answer ends up as x\ln{(2x+1)} but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
    For the integral -\int\frac{2x}{2x+1}\,dx, make the substitution z=2x+1\implies x=\tfrac{1}{2}\left(z-1\right)

    Thus, \,dz=2\,dx.

    The integral becomes: -\int\frac{x\,dz}{z}\implies -\tfrac{1}{2}\int \frac{z-1}{z}\,dz=\tfrac{1}{2}\int\left(\frac{1}{z}-1\right)\,dz

    Can you continue from here?
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  3. #3
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    Quote Originally Posted by stevedave View Post
    I'm having some trouble with this

    \int{\ln{(2x+1)}dx}

    I set u = ln(2x+1) and dv = dx so that du=\frac{2}{2x+1} and v = x.

    so after integrating by part once I get:

    x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}

    then I tried to integrate by part again with u=x and dv=(2x+1)^{-1}

    my final answer ends up as x\ln{(2x+1)} but my solution manual tells me differently.

    Can anyone tell me what I'm doing wrong?
    In the second integral, expand

    \int\frac{x}{2x+1}dx = \int \frac{1}{2} - \frac{1}{2}\frac{1}{2x+1} dx and integrate each term separately

    BTW  \int \frac{1}{ax+b} dx = \frac{1}{a} \ln| ax + b| + c
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  4. #4
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    \int \ln(2x+1) \, dx = x\ln(2x+1) - \int \frac{2x}{2x+1} \, dx<br />

    note that \frac{2x}{2x+1} = \frac{2x+1-1}{2x+1} = 1 - \frac{1}{2x+1}

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

    can you finish from here?
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  5. #5
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    Thank you everyone. That was very helpful.
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  6. #6
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    How about this for a short cut? Notice that one integral of 1 is x + 1/2, and make that the value of v, for dv/dx = 1 ...





    As usual, straight lines differentiate down / integrate up with respect to x.

    Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
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