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anonymous

  • 5 years ago

need help with evaluating this integral.... (x^9)/(sqrt(3+x^5)) dx.....

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  1. anonymous
    • 5 years ago
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    I would recommend a u substitution of u = x^5+3

  2. anonymous
    • 5 years ago
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    ok i started that....

  3. anonymous
    • 5 years ago
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    How far did you get?

  4. anonymous
    • 5 years ago
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    then i got (1/5)du = x^4dx ....

  5. anonymous
    • 5 years ago
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    Ok, though I prefer dx = du/(5x^4)

  6. anonymous
    • 5 years ago
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    ok.... then

  7. anonymous
    • 5 years ago
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    So then, what is x^5 in terms of u?

  8. anonymous
    • 5 years ago
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    i dont know what you mean by that

  9. anonymous
    • 5 years ago
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    u = x^5 + 3 so x^5 = ?

  10. anonymous
    • 5 years ago
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    x^5 = u - 3 right?

  11. anonymous
    • 5 years ago
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    u - 3

  12. anonymous
    • 5 years ago
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    ok. Good. Now start substituting in your integral

  13. anonymous
    • 5 years ago
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    so whats the x^9?... the denominator will be sqrt3?

  14. anonymous
    • 5 years ago
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    The denominator will be \(\sqrt{u}\)

  15. anonymous
    • 5 years ago
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    thats what i meant.///

  16. anonymous
    • 5 years ago
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    the dx will be \[\frac{1}{5x^4}du\]

  17. anonymous
    • 5 years ago
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    So what's left in the numerator( after you simplify)

  18. anonymous
    • 5 years ago
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    idk.......

  19. anonymous
    • 5 years ago
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    Cmon. \[\int \frac{x^9}{\sqrt{x^5+3}}dx = \int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx \] \[\text{Let }u = x^5+3 \]\[\implies x^5 = u-3 \] \[\implies 5x^4dx = du \implies dx = \frac{1}{5x^4}du\] Therefore \[\int \frac{x^5 * x^4}{\sqrt{x^5+3}}dx = \int \frac{(u-3)*x^4}{\sqrt{u}} (\frac{1}{5x^4})du\]

  20. anonymous
    • 5 years ago
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    And we can cancel the x^4 from top and bottom. Then split the fraction into two different fractions \[\frac{u}{\sqrt{u}} - \frac{3}{\sqrt{u}}\]

  21. anonymous
    • 5 years ago
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    and integrate each one separately.

  22. anonymous
    • 5 years ago
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    what happened to the 5 in the faraction

  23. anonymous
    • 5 years ago
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    yes there should be a factor of 5 outside the integral , he most likely didnt mention it because any integral of a constant multiple is just the integral times that constant

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