Evaluate $\int\dfrac{dx}{x^2-1}$ using partial fractions.

Question

Accepted Answer

**Factoring and decomposing**

$$x^2-1=(x-1)(x+1)$$

Let $\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}$, so $1=A(x+1)+B(x-1)$.

Setting $x=1$: $1=A(2)\Rightarrow A=\frac12$. Setting $x=-1$: $1=B(-2)\Rightarrow B=-\frac12$.

$$\frac{1}{x^2-1}=\frac{1}{2}\cdot\frac{1}{x-1}-\frac{1}{2}\cdot\frac{1}{x+1}$$

**Integrating term by term**

$$\int\frac{dx}{x^2-1}=\frac12\ln|x-1|-\frac12\ln|x+1|+C=\frac12\ln\left|\frac{x-1}{x+1}\right|+C$$ …

Q.Evaluate $\int\dfrac{dx}{x^2-1}$ using partial fractions.