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Math Help - basic integration

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    basic integration

    Hi, hope someone can help with this. Any working or explanation would be greatly appreciated. I really want to understand why/what I need to do/think. Thank you.

    Find the indefinite integral of dx/x^2 -9

    Cheers
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  2. #2
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    Quote Originally Posted by slaypullingcat View Post
    Hi, hope someone can help with this. Any working or explanation would be greatly appreciated. I really want to understand why/what I need to do/think. Thank you.

    Find the indefinite integral of dx/x^2 -9

    Cheers
    I assume you mean this

    \int \frac{dx}{x^2-9}

    The first thing to do is to factor the denominator and then use partial fraction decomp

    \frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}

    Clearing the fractions we get

    1=A(x+3)+B(x-3)

    if we let x=3 we get

    1=A(6) \iff A=\frac{1}{6}


    if we let x=-3 we get

    1=B(-6) \iff B=-\frac{1}{6}

    Now we have the integrals

    \int \frac{dx}{x^2-9}=\int \frac{\frac{1}{6}}{(x-3)}dx+\int \frac{-\frac{1}{6}}{(x+3)}dx

    The latter are both u subs for the denominator
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