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arsh125

  • 3 years ago

\[\int\limits_{1}^{6} x^2/\sqrt{x+3}\]

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  1. arsh125
    • 3 years ago
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    how do i evaluate that integral

  2. RadEn
    • 3 years ago
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    int by subs, let u=x+3

  3. UnkleRhaukus
    • 3 years ago
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    \[\int_1^6\frac{x^2}{\sqrt{x+3}}\text dx\] \(u=x+3\) \(\frac{\text du}{\text dx}=1\) \(\text du=\text dx\) \[=\int_1^6\frac{(u-3)^2}{\sqrt{u}}\text du\]

  4. UnkleRhaukus
    • 3 years ago
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    \[=\int_1^6\frac{u^2-6u+9}{\sqrt{u}}\text du\] \[=\int_1^6\frac{u^2}{\sqrt{u}}\text du-6\int_1^6\frac{u}{\sqrt{u}}\text du+9\int_1^6\frac{1}{\sqrt{u}}\text du\]

  5. lgbasallote
    • 3 years ago
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    ^?

  6. UnkleRhaukus
    • 3 years ago
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    no?

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