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Math Help - partial fractions

  1. #1
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    partial fractions

    <br />
\int\frac{dx}{1-x^2}<br />

    I am supposed to use partial fractions to solve this. I am pretty sure you could use a trig substitution here but how would you use partial fractions with this integral?
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  2. #2
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    \frac{1}{(1+x)(1-x)}=\frac{1}{2}\cdot \frac{(1+x)+(1-x)}{(1+x)(1-x)}=\frac{1}{2}\left( \frac{1}{1-x}+\frac{1}{1+x} \right).
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  3. #3
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    Quote Originally Posted by ur5pointos2slo View Post
    <br />
\int\frac{dx}{1-x^2}<br />

    I am supposed to use partial fractions to solve this. I am pretty sure you could use a trig substitution here but how would you use partial fractions with this integral?
    \frac{1}{1-x^2}=\frac{1}{(1+x)(1-x)}

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

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

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

    Now, in the numerator of both sides, equate the coefficients of constant terms, and x-terms,

    A + B =1

    A - B =0

    Solving we get, A = B = 1/2

    so, \int {\frac{1}{1-x^2}}~dx= \int {\left(\frac{1/2}{1+x}+\frac{1/2}{1-x}\right)}~dx

    =\frac{1}{2}\int{\left( \frac{1}{1+x}+\frac{1}{1-x}\right)}~dx
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