Evaluate the integral

Question

Evaluate the following indefinite integral:

∫(x4)/((x-3)*(x+4)) dx

Answer 1

Since the numerator is of a higher power than the denominator, it's easiest to first do long division.

 

(x-3)(x+4) = x^2 + x + 12

 

 

                          x^2 - x + 13

                         _____________________________

(x^2 + x - 12) / (x^4 + 0*x^3 + 0x^2 + 0*x + 0)

                     -(x^4  + 1*x^3 - 12*x^2 + 0*x + 0)

                                - 1*x^3 + 12*x^2 + 0*x + 0)

                             -(- 1*x^3 -   1*x^2 + 12*x + 0)

                                               13*x^2 - 12*x + 0

                                             -(13*x^2 +13*x - 156)

                                                             -25x +156

 

So the integral becomes:

 

∫x^2 - x + 13 + (-25x+156)/(x^2 + x - 12) dx

 

Break it up into two integrals.

 

∫x^2 - x + 13 dx + ∫(-25x+156)/(x^2 + x -12) dx

 

The first integral is easy, so I'll focus on the second on for now.

 

First change the denominator back to its original form.

 

∫(-25x+156)/((x+4)(x-3)) dx

 

Then do a partial fraction decomposition.

 

(-25x+156)/((x+4)(x-3)) = A/(x+4) + B/(x-3)

 

(-25x+156)/((x+4)(x-3)) = (A(x-3) + B(x+4))/((x+4)(x-3))

 

Cancel the denominators.

 

-25x + 156 = Ax - 3A + Bx + 4B

 

x(A+B) = -25x

-3A + 4B = 156

 

So

 

A + B = -25

-3A + 4B = 156

 

A = -B - 25

 

-3(-B - 25) + 4B = 156

3B + 75 + 4B = 156

7B = 81

B = 81/7

 

A + 81/7 = -175/7

A = -256/7

 

So we finally yield:

 

∫(x^2 - x + 13) dx + ∫(-256/7)/(x+4) dx + ∫(81/7)/(x-3) dx

 

Final answer:

 

1/3*x^3 - 1/2*x^2 + 13x + (-256/7)*ln(x+4) + (81/7)*ln(x-3) + C

 

Don't forget the +C!!!