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Math Help - I'm not sure how to approach this...

  1. #1
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    I'm not sure how to approach this...

    My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

    \int ln(2x+1) dx

    I can kind of get thorough it about halfway if i use f(x)=ln(2x+1) and g(x)=dx. but i end up with my second integral being this:

     \int \frac{2x}{2x+1}dx

    and i'm not sure how to get

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

    As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mortalapeman View Post
    My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

    \int ln(2x+1) dx

    I can kind of get thorough it about halfway if i use f(x)=ln(2x+1) and g(x)=dx. but i end up with my second integral being this:

     \int \frac{2x}{2x+1}dx

    and i'm not sure how to get

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

    As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.
     \int \frac{2x}{2x+1}dx=\int 1-\frac{1}{2x+1}~dx

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
     \int \frac{2x}{2x+1}dx=\int 1-\frac{1}{2x+1}~dx

    RonL
    What exactly do you mean? Are you saying that those two expressions are equal or that i need to substitute yours in for mine because i messed up somewhere along the way and that is the integral i should have got?

    Also, did you use the same substitutions as me?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mortalapeman View Post
    My problem is I don't i don't know what to substitute in for f(x) and g(x) when i try to use integration by parts for this particular integral.

    \int ln(2x+1) dx

    I can kind of get thorough it about halfway if i use f(x)=ln(2x+1) and g(x)=dx. but i end up with my second integral being this:

     \int \frac{2x}{2x+1}dx

    and i'm not sure how to get

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

    As my answer if i integrate that last one. I would post all my work but i'm kind of in a hurry. So thank you and I appreciate your help.
    \int\ln(2x+1)\,dx

    Let u=\ln(2x+1) and \,dv=\,dx. Thus \,du=\frac{2\,dx}{2x+1} and v=x.

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

    To integrate \int\frac{2x\,dx}{2x+1}, let u=2x+1\implies x=\frac{u-1}{2}

    Thus, \,du=2\,dx

    The integral is transformed into \frac{1}{2}\int\frac{u-1}{u}\,du=\frac{1}{2}\int\left[1-\frac{1}{u}\right]\,du=\frac{1}{2}u-\frac{1}{2}\ln|u|+C

    Resubstitute:

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

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

    \implies \left(x+\frac{1}{2}\right)\ln(2x+1)-\frac{1}{2}(2x+1)+C

    \implies \frac{1}{2}(2x+1)\ln(2x+1)-x\underbrace{-\frac{1}{2}+C}_{C}

    \implies \color{red}\boxed{\frac{1}{2}(2x+1)\ln(2x+1)-x+C}

    The \frac{1}{2} and C make up another constant, which I called C...

    I hope that this clarifies things!

    --Chris
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  5. #5
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    Quote Originally Posted by Chris L T521 View Post
    \int\ln(2x+1)\,dx

    Let u=\ln(2x+1) and \,dv=\,dx. Thus \,du=\frac{2\,dx}{2x+1} and v=x.

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

    To integrate \int\frac{2x\,dx}{2x+1}, let u=2x+1\implies x=\frac{u-1}{2}

    Thus, \,du=2\,dx

    The integral is transformed into \frac{1}{2}\int\frac{u-1}{u}\,du=\frac{1}{2}\int\left[1-\frac{1}{u}\right]\,du=\frac{1}{2}u-\frac{1}{2}\ln|u|+C

    Resubstitute:

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

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

    \implies \left(x+\frac{1}{2}\right)\ln(2x+1)-\frac{1}{2}(2x+1)+C

    \implies \frac{1}{2}(2x+1)\ln(2x+1)-x\underbrace{-\frac{1}{2}+C}_{C}

    \implies \color{red}\boxed{\frac{1}{2}(2x+1)\ln(2x+1)-x+C}

    The \frac{1}{2} and C make up another constant, which I called C...

    I hope that this clarifies things!

    --Chris
    Wow thank you so much, you have no idea >.< now i can work those other 10 problems just like that.
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