∫₁³(1/x²-16)

Question

Solve with definite integrals limits are given:

(3 and 1)

Answer 1

Use partial fraction decomposition.

 

1/(x²-16) = 1/[(x+4)(x-4)] = A/(x+4) + B/(x-4)

 

                1/[(x+4)(x-4)] = [A(x-4)+B(x+4)]/[(x+4)(x-4)]

 

                                  1 = A(x-4) + B(x+4)

 

                           0x+1 = (A+B)x+(-4A+4B)

 

So,   A + B = 0

     -4A +4B = 1

 

Solve for A and B to get A = -1/8 and B = 1/8.

 

The original integral is then equivalent to:

 

  (-1/8)∫_{(from 1 to 3)}(1/(x+4))dx + (1/8)∫_{(from 1 to 3)}(1/(x-4))dx

 

= (-1/8)lnlx+4l_{(from 1 to 3)} + (1/8)lnlx-4l_{(from 1 to 3)}

 

= (-1/8)[ln7-ln5] + (1/8)[ln1-ln3]

 

= (1/8)[ln5-ln7-ln3] = (1/8)[ln5-ln21] = (1/8)ln(5/21)