Partial Fractions — Question 5

Question 5

Find the value of $k$ that removes the $\ln|x+1|$ term. Then evaluate: $$\int\frac{kx+1}{x^2-1}\,dx.$$

Question 5 – Solution

Step 1: Set up the decomposition. $$\frac{kx+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.$$ Step 2: Clear the denominators. $$kx+1=A(x+1)+B(x-1).$$ Step 3: Find the coefficients. $$\begin{align*} x=1:&\quad k+1=2A \Longrightarrow A=\frac{k+1}{2},\\ x=-1:&\quad -k+1=-2B \Longrightarrow B=\frac{k-1}{2}. \end{align*}$$ Step 4: Remove the $\ln|x+1|$ term. Since integrating $B/(x+1)$ gives $B\ln|x+1|$, set $B=0$: $$B=0\quad\Longrightarrow\quad \frac{k-1}{2}=0 \quad\Longrightarrow\quad \boxed{k=1}.$$ Step 5: Substitute $k=1$, simplify, and integrate. $$\frac{x+1}{x^2-1}=\frac{x+1}{(x-1)(x+1)}=\frac1{x-1}, \qquad x\ne\pm1.$$ $$\begin{align*} \int\frac{x+1}{x^2-1}\,dx &=\int\frac{dx}{x-1}=\boxed{\ln|x-1|+C}. \end{align*}$$