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  1. #1
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    improper int

    how do u do this?
    int 0 to a (1/(a^2 - x^2))dx

    wut i did was to put c=0
    so lim c->0+ int c->a dx/(a^2 - x^2)

    am i doing this right?
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  2. #2
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    Quote Originally Posted by solars View Post
    how do u do this?
    int 0 to a (1/(a^2 - x^2))dx

    wut i did was to put c=0
    so lim c->0+ int c->a dx/(a^2 - x^2)

    am i doing this right?
     \int_0^a \frac{1}{x^2 - a^2} \, dx = \lim_{c \rightarrow a} \int_0^c \frac{1}{x^2 - a^2} \, dx .
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
     \int_0^a \frac{1}{x^2 - a^2} \, dx = \lim_{c \rightarrow a} \int_0^c \frac{1}{x^2 - a^2} \, dx .
    just one more question how do u go about integrating 1/(a^2-x^2)?
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  4. #4
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    Quote Originally Posted by solars View Post
    just one more question how do u go about integrating 1/(a^2-x^2)?
    Partial fraction decomposition.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by solars View Post
    just one more question how do u go about integrating 1/(a^2-x^2)?
    Here's an alternative to partial fractions :

    \frac{1}{a^2-x^2}=\frac{1}{(a-x)(a+x)}=\frac{1}{a}\left[\frac{a}{(a-x)(a+x)}\right]=\frac{1}{a}\left[\frac{a+x-x}{(a-x)(a+x)}\right] =\frac{1}{a}\left[\frac{a-x}{(a-x)(a+x)}+\frac{x}{a^2-x^2}\right]=\frac{1}{a}\left[\frac{1}{a+x}-\frac{x}{x^2-a^2}\right]

    So you would have \frac{1}{a}\left[\int\frac{\,dx}{x+a}-\int\frac{x\,dx}{x^2-a^2}\right]

    --Chris
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