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Math Help - [SOLVED] Improper Integral

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Improper Integral

    Last one!!

    \int\limits^{1}_{0} 35 \frac{ln(x)}{\sqrt x} dx

    Please don't give me the entire problem, I just can't figure out what technique I need to use to get started. Should I just do a substitution? I tried doing u=lnx, but that square root is throwing me.
    Thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Last one!!

    \int\limits^{1}_{0} 35 \frac{ln(x)}{\sqrt x} dx

    Please don't give me the entire problem, I just can't figure out what technique I need to use to get started. Should I just do a substitution? I tried doing u=lnx, but that square root is throwing me.
    Thanks!
    Your pick for the substitution is correct.

    Let u=\ln x\implies \,du=\frac{1}{x}\,dx

    The trick here is to note that 35\int_0^1\frac{\ln x}{\sqrt{x}}\,dx=35\int_0^1\frac{\sqrt{x}\ln x}{x}\,dx

    So now when you apply the substitution, the integral becomes 35\int_0^1 u\sqrt{x}\,du

    Now to get rid of \sqrt{x}, recall that we let u=\ln x\implies e^u=x\implies \sqrt{x}=e^{\frac{1}{2}u}

    Thus, you integral (including change of limits) becomes 35\int_{-\infty}^0 ue^{\frac{1}{2}u}\,du

    Can you take it from here?
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