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anonymous

  • 4 years ago

find the following indefinite integral using u-substitution to solve... HELP

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{}^{}6x \sqrt{x^2-25}dx\]

  2. anonymous
    • 4 years ago
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    I know I let u=x^2-25 which makes du=2xdx

  3. anonymous
    • 4 years ago
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    i am stuck at a particular step.. i will show you what i have so far

  4. mathmate
    • 4 years ago
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    hint: try u=sqrt(x^2-25) or try differentiating sqrt(x^2-25) and see what you get.

  5. dumbcow
    • 4 years ago
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    ok then dx = du/2x \[\int\limits_{}^{}\frac{6x \sqrt{u}}{2x} du =3 \int\limits_{}^{}\sqrt{u} du\]

  6. anonymous
    • 4 years ago
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    \[\int\limits_{}^{}\sqrt{x^2-25}\times6xdx = 3\int\limits_{}^{}\sqrt{x^2-25}\times2xdx = 3\int\limits_{}^{}\sqrt{u}du\]

  7. anonymous
    • 4 years ago
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    i'm lost after that

  8. dumbcow
    • 4 years ago
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    \[\sqrt{u} = u^{1/2}\] use the power rule

  9. dumbcow
    • 4 years ago
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    \[\large \int\limits_{}^{}x^{n} = \frac{x^{n+1}}{n+1}\]

  10. anonymous
    • 4 years ago
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    yeah so i would fill fill in x^2 -25 now for that step or no?

  11. dumbcow
    • 4 years ago
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    no leave it in terms of u until you have finished integrating, then the last step will be to substitute the x^2 -25 back in for u

  12. anonymous
    • 4 years ago
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    ok so i got 2u^(3/2)+c ... is that correct?

  13. dumbcow
    • 4 years ago
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    yep

  14. anonymous
    • 4 years ago
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    ok so then substitute in now? giving me.. 2(x^2-25)^(3/2) + c ..?

  15. dumbcow
    • 4 years ago
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    yes you could also write it as ...2sqrt(x^2-25)(x^2 -25) + c

  16. anonymous
    • 4 years ago
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    ok so then my final answer is \[2x^2-50\sqrt{x^2-25}+c\]..? or just leave it \[2(x^2-25)\sqrt{x^2-25}+c\]?

  17. dumbcow
    • 4 years ago
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    i would just leave it in factored form, but both are correct

  18. anonymous
    • 4 years ago
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    thank you so much!

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