Question

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

• A. $\frac{x^3}{3}-x+C$
• B. $\frac{x^5}{5}+x+C$
• C. $\frac{x^4}{4}+x+C$
• D. $\frac{x^2}{3}-x+C$

Answer 1

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

$\int x^2(1-\frac{1}{x^2})dx=\int (x^2-1)dx=\int x^{2}dx-\int 1dx=\frac{x^3}{3}-x+C$