Partial Fractions — Question 2

Question 2

Use polynomial division, then evaluate: $$\int\frac{x^3+1}{x^2-1}\,dx.$$ State the excluded values of $x$.

Question 2 – Solution

Step 1: Divide the polynomials. Since $x^3/x^2=x$, $$\begin{align*} x(x^2-1)&=x^3-x,\\ (x^3+1)-(x^3-x)&=x+1. \end{align*}$$ Thus $x^3+1=x(x^2-1)+(x+1)$, so $$\frac{x^3+1}{x^2-1}=x+\frac{x+1}{x^2-1}.$$ Step 2: Factor and simplify. $$\begin{align*} \frac{x^3+1}{x^2-1}&=x+\frac{x+1}{(x-1)(x+1)}\\ &=x+\frac1{x-1}, \end{align*}$$ because the original denominator is zero at both $x=-1$ and $x=1$.

Step 3: Integrate term by term. $$\begin{align*} \int\frac{x^3+1}{x^2-1}\,dx &=\int x\,dx+\int\frac{dx}{x-1}\\ &=\boxed{\frac{x^2}{2}+\ln|x-1|+C},\qquad x\ne\pm1. \end{align*}$$