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Math Help - Integration

  1. #1
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    Integration

    Hello all~ I'm doing a problem where Integration by parts is used and i have reached a part where i'm stumped.

    The problem is

    Integrate ln(2x+1)dx

    I've let u = ln(2x+1) dv = dx

    du = 2 / (2x+1)dx v = x


    xln(2x+1) - integrate (2x)/(2x + 1) dx

    I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into

    Integrate 1 - (1/(2x + 1)) which I can do fairly easily by using substitution. Any help is appreciated.
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  2. #2
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    Hello,

    I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into
    \frac{2x}{2x+1}=\frac{2x+1}{2x+1}-\frac{1}{2x+1}=1-\frac{1}{2x+1}

    Does it help ?
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  3. #3
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    Quote Originally Posted by JonathanEyoon View Post
    Hello all~ I'm doing a problem where Integration by parts is used and i have reached a part where i'm stumped.

    The problem is

    Integrate ln(2x+1)dx

    I've let u = ln(2x+1) dv = dx

    du = 2 / (2x+1)dx v = x


    xln(2x+1) - integrate (2x)/(2x + 1) dx

    I'm having problems integrating (2x) / (2x + 1) dx. It seems something so simple but when I tried it out a couple times, it simply doesn't match up to my teachers answer. He was able to make the above into

    Integrate 1 - (1/(2x + 1)) which I can do fairly easily by using substitution. Any help is appreciated.

    rewrite as follows (or use long division)

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

    I hope this helps.

    Edit: I'm too slow Go  {\mu}
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  4. #4
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    I'm sorry this is kinda embarrassing since it's algebra but could you tell me step by step why it's reduced like that? I remember vaguely but it was so long ago
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