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Math Help - How to integrate (x+1)/(2x+3)

  1. #1
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    How to integrate (x+1)/(2x+3)

    Hello,

    What is the method of integrating I = (x + 1)/(2x + 3)?

    I get a wrong answer: 0.5x - 0.5ln(2x + 3)
    The correct answer is 0.5x - 0.25ln(2x + 3)

    My method:

    Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

    (x+1) = A{2x + 3} + B{2}

    equating coefficient of x: 1 = 2A :. A = 0.5
    constants: 1 = 3A + B :. B = -0.5

    :. I = 0.5x - 0.5ln(2x + 3)

    Thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by algorithm View Post
    Hello,

    What is the method of integrating I = (x + 1)/(2x + 3)?

    I get a wrong answer: 0.5x - 0.5ln(2x + 3)
    The correct answer is 0.5x - 0.25ln(2x + 3)

    My method:

    Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

    (x+1) = A{2x + 3} + B{2}

    equating coefficient of x: 1 = 2A :. A = 0.5
    constants: 1 = 3A + B :. B = -0.5

    :. I = 0.5x - 0.5ln(2x + 3)

    Thank you
    several approaches here.

    you can do long division to find x + 1 \div 2x + 3 to get \frac 12 \left( 1 - \frac 1{2x + 3} \right) ...which should be easy to integrate

    or just use algebra:

    note that \frac {x + 1}{2x + 3} = \frac 12 \left( \frac {2x + 2}{2x + 3} \right) = \frac 12 \left( \frac {2x + 3 - 1}{2x + 3} \right) = \frac 12 \left(1 - \frac {1}{2x + 3} \right)
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  3. #3
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    It's also possible to substitute:
    \int\frac{x+1}{2x+3}\,\text{d}x=\left|\begin{array  }{rcl}2x+3&=&\varphi\\x&=&\dfrac{\varphi-3}{2}\\\text{d}x&=&\frac{1}{2}\,\text{d}\varphi\en  d{array}\right|=\frac{1}{4}\int\frac{\varphi-1}{\varphi}\,\text{d}\varphi=\frac{1}{4}\int\left(  1-\frac{1}{\varphi}\right)\,\text{d}\varphi
    So the solution is function F(x)=\frac{2x+3}{4}-\frac{1}{4}\ln|2x+3|+C
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by algorithm View Post
    My method:

    Let (x+1) = A{2x + 3} + B{differential coefficient of 2x + 3}

    (x+1) = A{2x + 3} + B{2}

    equating coefficient of x: 1 = 2A :. A = 0.5
    constants: 1 = 3A + B :. B = -0.5

    :. I = 0.5x - 0.5ln(2x + 3)
    The problem is that \ln(2x+3) is not an antiderivative of \frac{1}{2x+3}. Do you see why this is so ?
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  5. #5
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    Thanks for the replies.
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