Partial Fractions — Question 1

Question 1

Find $A$ and $B$, then evaluate the integral: $$\frac{7x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}.$$ $$\int\frac{7x+1}{(x-1)(x+2)}\,dx.$$

Question 1 – Solution

Step 1: Set up the decomposition. $$\frac{7x+1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}.$$ Step 2: Clear the denominators. Multiply by $(x-1)(x+2)$: $$7x+1=A(x+2)+B(x-1).$$ Step 3: Find the coefficients. $$\begin{align*} x=1:&\quad 8=3A \quad\Longrightarrow\quad A=\frac83,\\ x=-2:&\quad -13=-3B \quad\Longrightarrow\quad B=\frac{13}{3}. \end{align*}$$ Step 4: Substitute and integrate each term. $$\begin{align*} \int\frac{7x+1}{(x-1)(x+2)}\,dx &=\int\left(\frac{8/3}{x-1}+\frac{13/3}{x+2}\right)dx\\ &=\frac83\int\frac{dx}{x-1}+\frac{13}{3}\int\frac{dx}{x+2}\\ &=\boxed{\frac83\ln|x-1|+\frac{13}{3}\ln|x+2|+C}. \end{align*}$$