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Math Help - U substitution

  1. #1
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    U substitution

    Still confused with u substitution. Can someone show me how they would solve this by U substitution?

    the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

    u= ln(2x+3)
    du=2/2x+3 dx

    Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

    Thanks.
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  2. #2
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    Quote Originally Posted by mike21 View Post
    Still confused with u substitution. Can someone show me how they would solve this by U substitution?

    the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

    u= ln(2x+3)
    du=2/2x+3 dx

    Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

    Thanks.
    \int_{0.5}^1 \ln(2x+3) \cdot \frac{1}{2x+3} \, dx<br />

    u = \ln(2x+3)<br />

    du = \frac{2}{2x+3} \, dx<br />

    \frac{1}{2} \int_{0.5}^1 \ln(2x+3) \cdot \frac{2}{2x+3} \, dx

    substitute and reset limits ...

    \frac{1}{2} \int_{\ln{4}}^{\ln{5}} u \, du<br />

    finish it.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mike21 View Post
    Still confused with u substitution. Can someone show me how they would solve this by U substitution?

    the integral from .5 to 1 of ln(2x+3) over 2x+3 dx.

    u= ln(2x+3)
    du=2/2x+3 dx

    Where do i go from here? I was just puting the u on top so it would look like this... u/2x+3 but now i have a u and an x. What do I do?

    Thanks.
    \int_{1/2}^1\frac{\ln(2x+3)}{2x+3}\,dx

    You are correct that

    u= \ln(2x+3)
    du=\frac{2}{2x+3}\,dx

    Notice that \frac{1}{2}u\,du is the original expression.

    So you have \int\frac{1}{2}u\,du

    To determine the bounds, simply plug them into the equation for u:

    Lower is \ln(2\cdot0.5 + 3) = \ln4
    Upper is \ln(2\cdot1+3)=\ln5

    So the integral that you need to evaluate is \frac{1}{2}\int_{\ln4}^{\ln5}u\,du
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