Question

$\int \frac{x^3-x^2+x-1}{x-1}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{x^3}{3}+x+C$
• B. $\frac{x^2}{2}+x+C$
• C. $\frac{x^4}{4}+x+C$
• D. $\frac{x^2}{2}-x+C$

Answer 1

The correct option is A $\frac{x^3}{3}+x+C$

$\int \frac{x^3-x^2+x-1}{x-1}dx=\int \frac{x^2(x-1)+x-1}{x-1}dx=\int (x^2+1)dx$
By taking the terms separately, we get
$=\int x^{2}dx+\int 1dx$
Therefore, we obtain
$=\frac{x^3}{3}+x+C$