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Math Help - Can't figure out how to find the integral of this equation

  1. #1
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    Can't figure out how to find the integral of this equation

    how would I solve the integral of 1/(a+b*sqrt(y))dy

    i've tried various approaches but I can't seem to get anywhere. Anyone have any strategies or solutions? (Also checked quite a few integral tables, can't find it anywhere)

    also the value at y=0 should be 0.
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  2. #2
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    Go to this website W|A
    In the input window, type in this exact phrase: integrate 1/[a +b sqrt[y]]dy.
    Click the equals bar at the end if the input window.
    Be sure you click ‘show steps’ to see the solution.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by boiler1987 View Post
    how would I solve the integral of 1/(a+b*sqrt(y))dy

    i've tried various approaches but I can't seem to get anywhere. Anyone have any strategies or solutions? (Also checked quite a few integral tables, can't find it anywhere)

    also the value at y=0 should be 0.
    \int {\frac{{dy}}{{a + b\sqrt y }}}  = \left\{ \begin{gathered}y = {t^2}, \hfill \\dy = 2tdt \hfill \\ \end{gathered}  \right\} = 2\int {\frac{t}{{a + bt}}} dt =

    = \frac{2}{b}\int {\frac{{a + bt - a}}{{a + bt}}} dt = \frac{2}{b}\int {\left( {1 - \frac{a}{{a + bt}}} \right)} dt =

    = \frac{2}{b}t - \frac{{2a}}{b}\int {\frac{{dt}}{{a + bt}}}  = \frac{2}{b}t - \frac{{2a}}{{{b^2}}}\int {\frac{{d\left( {a + bt} \right)}}{{a + bt}}}  =

    = \frac{2}{b}t - \frac{{2a}}{{{b^2}}}\ln \left| {a + bt} \right| + C = \frac{2}{b}\sqrt y  - \frac{{2a}}{{{b^2}}}\ln \left| {a + b\sqrt y } \right| + C.
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