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Math Help - [SOLVED] integral evaluation

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    [SOLVED] integral evaluation

    use partial fractions to derive the integration formula
    \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln\frac{a+x}{a-x}+c
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  2. #2
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    Quote Originally Posted by vinson24 View Post
    use partial fractions to derive the integration formula
    \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}ln\frac{a+x}{a-x}+c
    \frac{1}{a^2 - x^2} = \frac{1}{(a - x)(a + x)} = \frac{A}{a - x} + \frac{B}{a + x}.

    Cross multiplying gives

    \frac{1}{a^2 - x^2} = \frac{A(a + x) + B(a - x)}{a^2 - x^2}

    So A(a + x) + B(a - x) = 1

    Aa + Ax + Ba - Bx = 1

    (A - B)x + (A + B)a = 0x + 1.

    Equating like powers of x gives

    A - B = 0 and (A + B)a = 1.

    Solving these simultaneously gives A = B and therefore (A + B) = \frac{1}{a}

    2A = \frac{1}{a}

    A = \frac{1}{2a}.


    Therefore A = B = \frac{1}{2a}.


    So the integral becomes

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

     = -\frac{1}{2a}\ln{(a - x)} + \frac{1}{2a}\ln{(a + x)} + C

     = \frac{1}{2a}[\ln{(a + x)} - \ln{(a - x)}] + C

     = \frac{1}{2a}\ln{\left(\frac{a + x}{a - x}\right)} + C.
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