Post a New Question

calculus

posted by .

what is the integral of (4x^2-2)/(x^2-6x-40)

  • calculus -

    First do a long division
    (4x^2-2)/(x^2-6x-40) = 4 + (24x+158)/(x^2 - 6x - 40)

    now separate (24x+158)/(x^2 - 6x - 40
    into partial fractions so that
    A/(x-10) + B(x+4) = (24x + 158)/((x-10)(x+4))

    then
    A(x+4) + B(x-10) = 24x + 158
    let x = -4
    -14B = 62
    B = -31/7

    let x = 10
    14A = 398
    A = 199/7

    (4x^2-2)/(x^2-6x-40) = 4 + (199/7)/(x-10) - (31/7)/(x+4)

    so the integral is
    4x + (199/7)ln(x-10) - (31/7)ln(x+10) + C , where C is a constant

  • correction - calculus -

    I have a typo, the last line should say:


    4x + (199/7)ln(x-10) - (31/7)ln(x+4) + C , where C is a constant

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question