Question:
Evaluate: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
Solution:
Given, $\int \frac{x^{2}-x^{2}+x-1}{x-1} d x$.
$=\int \frac{x^{2}(x-1)+x-1}{x-1} d x$
$=\int \frac{(x-1)\left[x^{2}+1\right]}{x-1} d x$
$=\int\left(x^{2}+1\right) d x\left[\right.$ since, $\left.\int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$
$=\frac{x^{3}}{3}+x+c$