Partial Fractions — Question 3

Question 3

Decompose into partial fractions. Include every power of $(x-1)$: $$\frac{2x+3}{x(x-1)^2}.$$ Do not integrate.

Question 3 – Solution

Step 1: Include a term for every factor and repeated power. $$\frac{2x+3}{x(x-1)^2} =\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}.$$ Step 2: Clear the denominators. $$2x+3=A(x-1)^2+Bx(x-1)+Cx.$$ Step 3: Find $A$ and $C$. $$\begin{align*} x=0:&\quad 3=A(-1)^2 \Longrightarrow A=3,\\ x=1:&\quad 5=C(1) \Longrightarrow C=5. \end{align*}$$ Step 4: Find $B$ by using $x=2$. $$\begin{align*} 7&=3(1)^2+B(2)(1)+5(2)\\ &=3+2B+10,\\ -6&=2B \Longrightarrow B=-3. \end{align*}$$ Step 5: State the result. $$\boxed{\displaystyle \frac{2x+3}{x(x-1)^2}=\frac3x-\frac3{x-1}+\frac5{(x-1)^2}},$$ and the requested coefficients are $$\boxed{A=3,\qquad B=-3,\qquad C=5}.$$