Calculus integration

Evaluate the integral.

integeal {dx} / {(x-6)(x-3)(x+1)}

∫dx/[(x - 6)(x - 3)(x + 1)]

Using partial fractions:

A/(x - 6) + B/(x - 3) + C/(x + 1) = 1/[(x - 6)(x - 3)(x + 1)]

A(x - 3)(x + 1) + B(x - 6)(x + 1) + C(x - 6)(x - 3) = 1

When x = 3:

-12B = 1 => B = -1/12

When x = -1:

28C = 1 => C = 1/28

When x = 6:

21A = 1 => A = 1/21

The integral therefore becomes:

∫1/21*1/(x - 6) - 1/12*1/(x - 3) + 1/28*1/(x + 1) dx

= 1/21*ln|x - 6| - 1/12*ln|x - 3| + 1/28*ln|x + 1| + C