Partial Fractions — Question 9

Question 9

Use partial fractions to evaluate: $$\int\frac{dx}{x^3-x^2} =\int\frac{dx}{x^2(x-1)}.$$

Question 9 – Solution

Step 1: Set up partial fractions. The repeated factor $x^2$ requires both $1/x$ and $1/x^2$: $$\frac1{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}.$$ Step 2: Clear the denominators. $$1=Ax(x-1)+B(x-1)+Cx^2.$$ Step 3: Find $B$ and $C$. $$\begin{align*} x=0:&\quad 1=-B \Longrightarrow B=-1,\\ x=1:&\quad 1=C \Longrightarrow C=1. \end{align*}$$ Step 4: Find $A$ by setting $x=2$. $$\begin{align*} 1&=A(2)(1)+(-1)(1)+(1)(4)\\ &=2A+3,\\ -2&=2A \Longrightarrow A=-1. \end{align*}$$ Step 5: Substitute and integrate each term. $$\begin{align*} I&=\int\left(-\frac1x-\frac1{x^2}+\frac1{x-1}\right)dx\\ &=-\int\frac{dx}{x}-\int x^{-2}\,dx+\int\frac{dx}{x-1}\\ &=-\ln|x|-\frac{x^{-1}}{-1}+\ln|x-1|+C\\ &=-\ln|x|+\frac1x+\ln|x-1|+C. \end{align*}$$ Therefore, $$\boxed{\displaystyle I=-\ln|x|+\frac1x+\ln|x-1|+C}.$$