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anonymous

  • 4 years ago

integrate x^33^x dx

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  1. anonymous
    • 4 years ago
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    product of x raised to power three and 3 raised to power x

  2. anonymous
    • 4 years ago
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    i assume this is \[\int x^33^xdx\]

  3. anonymous
    • 4 years ago
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    ya

  4. anonymous
    • 4 years ago
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    it is a set up for integration by parts, but you have to do it 3 times ! will reduce the power on \[x^3\] once each time

  5. anonymous
    • 4 years ago
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    i have assumed f(x)=x^3 and g'(x)=3^x

  6. anonymous
    • 4 years ago
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    for example in the first time you will put \[u=x^3, dv=3^xdx, du=3x^2,v=\frac{3^x}{\ln(3)}\] and get \[uv-\int vdu=\frac{x^33^x}{\ln(3)}-\frac{3}{\ln(3)}\int x^23^xdx\]

  7. anonymous
    • 4 years ago
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    then lather, rinse, repeat. hold on

  8. anonymous
    • 4 years ago
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    easier to read it , click "show steps" http://www.wolframalpha.com/input/?i=x^3*3^x+dx

  9. anonymous
    • 4 years ago
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    how would u integrate 3x^2.3^x/log3

  10. anonymous
    • 4 years ago
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    first of all pull out the constants like i did above and get \[-\frac{3}{\ln(3)}\int x^23^xdx\] then repeat the parts, this time with \[u=x^2\]

  11. anonymous
    • 4 years ago
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    ur ans went over my head

  12. anonymous
    • 4 years ago
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    ok it is confusing, but if you got the first one right you should have a good idea what is going on each time you integrate by parts, you will reduce the power by one when you take the derivative of \[x^3, x^2, x\]

  13. anonymous
    • 4 years ago
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    don't forget that \[\ln(3)\] is a number (constant) so it come right out front of the integral sign

  14. anonymous
    • 4 years ago
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    your next job is really \[\int x^2 3^xdx\] the constant just sits out front. integrate by parts again

  15. anonymous
    • 4 years ago
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    let me show u the first step......3x^2.3x/log3-integrate3x^2.3^x/log3 dx

  16. anonymous
    • 4 years ago
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    ah first step is wrong that is the problem

  17. anonymous
    • 4 years ago
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    it is \[\int udv=uv-\int vdu\]

  18. anonymous
    • 4 years ago
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    i dnt use that formula

  19. anonymous
    • 4 years ago
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    \[u=x^3,du = 3x^2, dv = 3^x, v = \int 3^x=\frac{3^x}{\ln(3)}\]

  20. anonymous
    • 4 years ago
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    \[uv=\frac{x^33^x}{\ln(3)}\]

  21. anonymous
    • 4 years ago
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    and i stress again, even though the second one is \[-\int \frac{3x^23^x}{\ln(3)}dx\] you should rewrite it immediately as \[-\frac{3}{\ln(3)}\int x^23^xdx\]

  22. anonymous
    • 4 years ago
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    i m having trouble with the formula implementation......wht to do?

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