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 $1=A(x-3)+B(x+1)=(A+B)x+B-3A$, as an equation that holds for all $x$. Therefore, we must have

$${\begin{cases}A+B&=0\\B-3A&=1\end{cases}}$$

From the first equation, $B=-A$. Substituting this into the second equation yields $-4A=1$, so we have $-B=A={\frac {-1}{4}}$. We can therefore rewrite and solve the antiderivative:

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