Question

Use partial fractions to find the indefinite integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

∫x2/x1−20x2−125dx

Answer 1

125/6(In(x-25)) - 5/6(In(x+5))+C

Explanation:

∫x2/x1−20x2−125dx

Should be

∫x²/(x²−20x−125)dx

First of all let's factorize the denominator.

x²−20x−125= x²+5x-25x-125

x²−20x−125= x(x+5) -25(x+5)

x²−20x−125= (x-25)(x+5)

∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx

x²/(x²−20x−125) =x²/((x-25)(x+5))

x²/((x-25)(x+5))= a/(x-25) +b/(x+5)

x²/= a(x+5) + b(x-25)

Let x=25

625 = a30

a= 625/30

a= 125/6

Let x= -5

25 = -30b

b= 25/-30

b= -5/6

x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)

∫x²/(x²−20x−125)dx

=∫125/6(x-25) -∫5/6(x+5) Dx

= 125/6(In(x-25)) - 5/6(In(x+5))+C