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anonymous

  • 5 years ago

Find the integral:

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  1. anonymous
    • 5 years ago
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    \[\sqrt{x} + 1/ x^4 dx\]

  2. anonymous
    • 5 years ago
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    what is the integral of sqrt x

  3. anonymous
    • 5 years ago
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    that also is \[(x)^{1/2} + (x)^{-4}\]

  4. anonymous
    • 5 years ago
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    You can rewrite that integral as the integral of (x^(1/2) + x^(-4)) Using the power rule, we get: (x^(3/2) / (3/2)) + (x^(-3) / -3) which may be rewritten as (2/3) x^(3/2) - 3/x^3

  5. anonymous
    • 5 years ago
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    just integrate each part individually then add together

  6. anonymous
    • 5 years ago
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    add 1 to the exponents and w/e you get multiply each part by it's inverse

  7. anonymous
    • 5 years ago
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    you can't add them. If he could then he would do it before the integration

  8. anonymous
    • 5 years ago
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    integral of sqrt x = [2x^(3/2)]/3 integral of 1/x^4= -1/3x^3 Then add together and you get (2 x^(9/2)-1)/(3 x^3)+ C

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