Posted by **Anonymous** on Monday, October 3, 2011 at 8:25pm.

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

- calculus -
**Reiny**, Monday, October 3, 2011 at 9:01pm
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 -
**Reiny**, Monday, October 3, 2011 at 9:12pm
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

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