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Math Help - Please help me to find the integrals to these two problems

  1. #1
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    Please help me to find the integrals to these two problems

    ∫2-3x/9x^2+25 DX


    ∫( 1 on the top and 0 on the bottom of this symbol) x^2+2x-5/x+3 DX

    Please guys I really dont understand this. I would appreciate it if you can explain to me step by step on how I would solve these. Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    Break this up into...

    2\int{dx\over 9x^2+25} -3\int{xdx\over 9x^2+25}

    The first is an arctangent and the second is a basic sub with u=9x^2+25

    The second integral is a partial fractions, where you should start by dividing.
    Last edited by matheagle; May 10th 2010 at 03:07 PM.
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  3. #3
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    \int_0^1 \frac{x^2+ 2x-5}{x+3} dx

    Since the denominator has lower degree than the numerator, start by dividing: \frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}

    \int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}
    For that last integral, make the substitution u= x+ 3.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    \int_0^1 \frac{x^2+ 2x-5}{x+3} dx

    Since the denominator has lower degree than the numerator, start by dividing: \frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}

    \int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}
    For that last integral, make the substitution u= x+ 3.
    you have a typo in the upper limit in the first integral
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    \int_0^1 \frac{x^2+ 2x-5}{x+3} dx

    Since the denominator has lower degree than the numerator, start by dividing: \frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}

    \int_0^1 (x+ 5+ \frac{10}{x+ 3}dx= \int_0^5 (x+ 5) dx+ 10\int_0^1 \frac{dx}{x+ 3}
    For that last integral, make the substitution u= x+ 3.
    I factor the numerator and obtain a different result.

     \frac{x^2+ 2x-5}{x+3} = \frac{x(x+2) - 5}{x+3} = \frac{x(x+2+1-1) - 5}{x+3} = \frac{ x(x+3) - x - 5 }{x+3}

     x - \frac{x+5}{x+3} = x - \frac{(x+3)+2}{x+3} = x - 1 - \frac{2}{x+3}

    I cant see where I've made a mistake?

    Edit- I subbed in numbers and found that yours has an error somewhere

    \frac{x^2+ 2x- 5}{x+ 3}= x+ 5+ \frac{10}{x+ 3}

    Using x=2

    \frac{4+ 4 - 5}{2+ 3}= 2+ 5+ \frac{10}{2+ 3}

    \frac{3}{5}= 2+ 5+ \frac{10}{5} = 9 which is not true obviously

    Edit Edit (lol) -

    The second you is a partial fractions, where you should start by dividing.
    I hate dividing polynomials, if you can I would suggest factoring which can be done in this case. Obviously not everybody will share my hatred for dividing polynomials but I treat them like the annoying cousin: when I have to go see them I will, but not a day before!
    Last edited by AllanCuz; May 10th 2010 at 09:40 AM.
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