Let $f(x)=\frac{x^2-1}{x-1}$ for $x\ne 1$ and $f(1)=4$. What is the value of $\int_0^2 f(x)\,dx$?

Core Concept

For $x\ne 1$, $f(x)=x+1$, and changing the value at one point does not affect the integral. Then $\int_0^2(x+1)\,dx=[x^2/2+x]_0^2=2+2=4$.