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Math Help - Indefinite integral

  1. #1
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    Indefinite integral

    1) ∫(7x^(3) - 4x^(2) +9x -10)dx
    2) ∫ [(6 √(x)+2)/ √(x)]
    3) ∫6x^(7) +4 / (x)

    Thanks
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  2. #2
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    Dear bgwarhawk,

    these are elementary intagrals. You must recall the formula:

    \int x^a dx = x^{a+1}/a+C (if a <> -1)
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    Quote Originally Posted by Skalkaz View Post
    Dear bgwarhawk,

    these are elementary intagrals. You must recall the formula:

    \int x^a dx = x^{a+1}/a+C (if a <> -1)
    What in the world is that?
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  4. #4
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    This is a formula from second page in a book about integral. What is your problem? Where did you get stuck? Or do you think this forum is a automatic solution generator of homework?
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    Quote Originally Posted by Skalkaz View Post
    This is a formula from second page in a book about integral. What is your problem? Where did you get stuck? Or do you think this forum is a automatic solution generator of homework?
    For the 1) I got 7/2x^-4/3x^3+9/2x^2-10x but thats not right...
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  6. #6
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    the 1st term is incorrect. And don't forget the C at the end.
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  7. #7
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    Quote Originally Posted by bgwarhawk View Post
    1) ∫(7x^(3) - 4x^(2) +9x -10)dx
    2) ∫ [(6 √(x)+2)/ √(x)]
    3) ∫6x^(7) +4 / (x)

    Thanks
    1) ∫(7x^(3) - 4x^(2) +9x -10)dx=  \frac{7x^4}{4}-\frac{4x^3}{3}+\frac{9x^2}{2}-10+C

    2) divide \sqrt{x} into both terms

    3)power rule and natural log
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  8. #8
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    Quote Originally Posted by Skalkaz View Post
    Dear bgwarhawk,

    these are elementary intagrals. You must recall the formula:

    \int x^a dx = x^{a+1}/a+C (if a <> -1)
    Actually it's \int x^a dx = \frac{1}{a + 1}x^{a+1}+C.
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  9. #9
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    Quote Originally Posted by Erdos32212 View Post
    1) ∫(7x^(3) - 4x^(2) +9x -10)dx=  \frac{7x^4}{4}-\frac{4x^3}{3}+\frac{9x^2}{2}-10+C

    2) divide \sqrt{x} into both terms

    3)power rule and natural log
    1. \int{7x^3 - 4x^2 + 9x - 10\,dx} = \frac{7}{4}x^4 - \frac{4}{3}x^3 + \frac{9}{2}x^2 - 10x + C
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  10. #10
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    Quote Originally Posted by bgwarhawk View Post
    1) ∫(7x^(3) - 4x^(2) +9x -10)dx
    2) ∫ [(6 √(x)+2)/ √(x)]
    3) ∫6x^(7) +4 / (x)

    Thanks
    2. \int{\frac{6\sqrt{x} +2}{\sqrt{x}}\,dx} = \int{6 + \frac{2}{\sqrt{x}}\,dx} = \int{6 + 2x^{-\frac{1}{2}}\,dx}

    Can you go from there?
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  11. #11
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    Quote Originally Posted by Prove It View Post
    2. \int{\frac{6\sqrt{x} +2}{\sqrt{x}}\,dx} = \int{6 + \frac{2}{\sqrt{x}}\,dx} = \int{6 + 2x^{-\frac{1}{2}}\,dx}

    Can you go from there?
    What do I do?
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  12. #12
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    Quote Originally Posted by bgwarhawk View Post
    What do I do?
    Use the formula \int x^a dx = \frac{1}{a + 1}x^{a+1}+C.

    \int{6 + 2x^{-\frac{1}{2}}\,dx} = \int{6x^0 + 2x^{-\frac{1}{2}}\,dx}

     = \frac{6}{1}x^1 + \frac{2}{\frac{1}{2}}x^{\frac{1}{2}} + C

     = 6x + 4x^{\frac{1}{2}} + C

     = 6x + 4\sqrt{x} +C.
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