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KatClaire

  • 3 years ago

Find the integral!!!

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  1. KatClaire
    • 3 years ago
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    \[\int\limits_{}^{} \frac{ 3 }{ \sqrt[3]{3x-5} } dx\]

  2. KatClaire
    • 3 years ago
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    I really don't have much a clue for this one

  3. Yahoo!
    • 3 years ago
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    Let 3x - 5 = u

  4. Yahoo!
    • 3 years ago
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    Now..Use Substitution Method

  5. KatClaire
    • 3 years ago
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    and I take out the 3 on top right?

  6. Mimi_x3
    • 3 years ago
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    \(u=3x+5\) du/dx=3 \[\int\limits\frac{3}{\sqrt[3]{u}} *\frac{du}{3} =>\int\limits\frac{1}{\sqrt[3]{u}} du=>\int\limits\frac{1}{u^{1/3}} du\]

  7. Yahoo!
    • 3 years ago
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    du / dx = 3 du = 3 dx

  8. Mimi_x3
    • 3 years ago
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    u=3x-5**

  9. KatClaire
    • 3 years ago
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    perfect! I'll try it again

  10. KatClaire
    • 3 years ago
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    would it be \[\frac{ 3(3x-5)^{2/3} }{ 2 } +c\]

  11. Ravi2
    • 3 years ago
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    is this method called integration by substitution method??

  12. KatClaire
    • 3 years ago
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    I think so

  13. sirm3d
    • 3 years ago
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    you can also try \[y=\sqrt[3]{3x-5}\Rightarrow y^3=3x-5\\3y^2\;\mathrm dy =3\;\mathrm dx\]

  14. Ravi2
    • 3 years ago
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    yup.....also an other way is by doing direct inverse power......

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