Evaluate the following integral:
\[
\int \frac{1}{(x-1)(x+2)} d x
\]
a. $\frac{1}{3} \ln \left|\frac{x+2}{1-x}\right|+C$
b. $\ln \left|\frac{1-x}{x+2}\right|+C$
c. $\frac{1}{3} \ln \left|\frac{1-x}{x+2}\right|+C$
d. $\ln |(1-x)(x+2)|+C$

Step 1 ：Given the integral \(\int \frac{1}{(x-1)(x+2)} dx\), we can solve it by using the method of partial fractions.

Step 2 ：First, we decompose the integrand into partial fractions. The integrand is \(\frac{1}{(x-1)(x+2)}\), which can be decomposed into \(\frac{A}{x-1} + \frac{B}{x+2}\), where A and B are constants to be determined.

Step 3 ：Solving for A and B, we find that A = \(\frac{1}{3}\) and B = -\(\frac{1}{3}\).

Step 4 ：Now, we can integrate each term separately. The integral of \(\frac{1}{3} \over x-1\) is \(\frac{1}{3} \ln |x-1|\), and the integral of -\(\frac{1}{3} \over x+2\) is -\(\frac{1}{3} \ln |x+2|\).

Step 5 ：Adding the two integrals together, we get \(\frac{1}{3} \ln |x-1| - \frac{1}{3} \ln |x+2|\).

Step 6 ：Using the properties of logarithms, we can simplify this to \(\frac{1}{3} \ln \left|\frac{x-1}{x+2}\right|\).

Step 7 ：Finally, we add the constant of integration C to get the final answer: \(\boxed{\frac{1}{3} \ln \left|\frac{x-1}{x+2}\right| + C}\). The correct option is (a).