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Math Help - hard integration problem

  1. #1
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    hard integration problem

    problem is integrate (5^x + 2e^5lnx) dx

    I can only think to take the natural log of both sides :

    xln5 + (5lnx)(ln2e)

    so then I try to integrate by parts for xln5 and get :
    xln5 = (ln5(x^2))/2 - (ln5(x^2))/2 + xln5 OR 0=0.

    So integration by parts for xln5 wont work

    So next thing and last thing I can think of is to combine ln5^x +ln2e^5lnx =
    integral of ln[(5^x)(2e^5lnx)]

    But how do I do this? no clue.
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  2. #2
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    Is it 2e^(5lnx)?
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  3. #3
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    Quote Originally Posted by frankinaround View Post
    problem is integrate (5^x + 2e^5lnx) dx

    I can only think to take the natural log of both sides :

    xln5 + (5lnx)(ln2e)

    so then I try to integrate by parts for xln5 and get :
    xln5 = (ln5(x^2))/2 - (ln5(x^2))/2 + xln5 OR 0=0.

    So integration by parts for xln5 wont work

    So next thing and last thing I can think of is to combine ln5^x +ln2e^5lnx =
    integral of ln[(5^x)(2e^5lnx)]

    But how do I do this? no clue.
    e^{5 \ln x} = e^{\ln x^5} = x^5

    5^x = e^{kx} where k = \ln (5).

    You should know how to integrate both of the above.
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    e^{5 \ln x} = e^{\ln x^5} = x^5

    5^x = e^{kx} where k = \ln (5).

    You should know how to integrate both of the above.
    wow wow yeah I get it. 2e^5lnx = 2x^5 right, but 5^x I dont follow AS well, can you elaborate on that just a LITTLE bit?
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  5. #5
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    integral of a^x is (a^x)/(ln a) + C
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  6. #6
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    Quote Originally Posted by frankinaround View Post
    wow wow yeah I get it. 2e^5lnx = 2x^5 right, but 5^x I dont follow AS well, can you elaborate on that just a LITTLE bit?
    Let y = 5^x.

    Then ln(y) = x ln(5).

    Exponentiate both sides to base e:

    y = 5^x = e^{x ln(5)}.
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