Posted by Anonymous on Monday, October 3, 2011 at 8:25pm.
what is the integral of (4x^22)/(x^26x40)

calculus  Reiny, Monday, October 3, 2011 at 9:01pm
First do a long division
(4x^22)/(x^26x40) = 4 + (24x+158)/(x^2  6x  40)
now separate (24x+158)/(x^2  6x  40
into partial fractions so that
A/(x10) + B(x+4) = (24x + 158)/((x10)(x+4))
then
A(x+4) + B(x10) = 24x + 158
let x = 4
14B = 62
B = 31/7
let x = 10
14A = 398
A = 199/7
(4x^22)/(x^26x40) = 4 + (199/7)/(x10)  (31/7)/(x+4)
so the integral is
4x + (199/7)ln(x10)  (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(x10)  (31/7)ln(x+4) + C , where C is a constant
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