how to solve this question ∫dx/x2-9 ?

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  • Anonymous
    9 years ago
    Favorite Answer

    Using partial fractions:

    1/(x²-9)=A/(x+3) + B/(x-3).....(i)

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

    So, you get A=1/6 and B= -1/6

    Now:substituting back A and B in (i)

    1/(x²-9) = -1/ 6(x+3) + 1/ 6(x-3)

    ∫dx/x2-9

    =∫ -1/ 6(x+3) + 1/ 6(x-3) dx

    =∫ -1/ 6(x+3) dx + ∫ 1/ 6(x-3) dx

    = -1/ 6 ∫ 1/(x+3) dx + 1/ 6 ∫ 1/(x-3) dx

    = -1/ 6 ln|x+3| + 1/ 6 ln |x-3| + C

    = -1/ 6 ln|x+3| + 1/ 6 ln |x-3| + C

    = 1/ 6 [ ln|x-3| - ln |x+3|] + C

    = 1/ 6 ln |(x-3)/(x+3)| + C..........that is ya answer

  • 9 years ago

    Is that ∫ dx/(x² - 9) ?

    If so, then

    ∫ dx/(x² - 9) = (-1/3) arctanh(x/3) + C.

    Alternatively, you can use a partial fraction decomposition

    1/(x² - 9) = (1/6)/(x - 3) - (1/6)/(x + 3).

    So

    ∫ dx/(x² - 9) = (1/6) ln|(x-3)/(x+3)| + C

    The answers are the same.

  • Como
    Lv 7
    9 years ago

    Presentation looks dodgy.

    Will take a guess that is meant to read as :-

    I = ∫ dx / (x² - 9) = ∫ dx / [ (x - 3) (x + 3) ]

    1 / [ (x - 3)(x + 3) ] = A / (x - 3) + B / (x + 3)

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

    A = 1/6

    B = -1/6

    I = (1/6) ∫ dx / (x - 3) - (1/6) ∫ dx / (x + 3)

    I = (1/6) log (x - 3) - (1/6) log (x + 3) + C

  • Mark
    Lv 7
    9 years ago

    I guess you mean ∫ dx/(x²–9)

    Partial fractions is the most direct way. (You can also use inverse hyperbolic functions if you know the formulas, but this is the more elementary method.)

    1/(x²–9) = 1/(x+3)(x–3) = (–1/6)/(x+3) + (1/6)/(x–3)

    So we have (1/6) [ ∫ dx/(x–3) – ∫ dx/(x+3) ] =

    (1/6) [ ln |x–3| – ln |x + 3| ] + constant

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