Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 15/Solution 1

We want to find $A, B$ so that $\frac{1}{(x + 1) (x - 3)} = \frac{A}{x + 1} + \frac{B}{x - 3} .$ Since the right-hand side is equal to $\frac{A (x - 3) + B (x + 1)}{(x + 1) (x - 3)}$ , we must have $1 = A (x - 3) + B (x + 1) = (A + B) x + B - 3 A$ , as an equation that holds for all $x$ . Therefore, we must have ${\begin{cases} A + B & = 0 \\ B - 3 A & = 1 \end{cases}$ From the first equation, $B = - A$ . Substituting this into the second equation yields $- 4 A = 1$ , so we have $- B = A = \frac{- 1}{4}$ . We can therefore rewrite and solve the antiderivative:

$\int \frac{1}{(x + 1) (x - 3)} d x = \int (\frac{- 1 / 4}{x + 1} + \frac{1 / 4}{x + 3}) d x = \frac{- 1}{4} \ln | x + 1 | + \frac{1}{4} \ln | x - 3 | + C = \frac{1}{4} \ln | \frac{x - 3}{x + 1} | + C .$