Question

The value of $\int (3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$ is

• A. $x^3\tan \frac{1}{x}+c$
• B. $x^2\tan \frac{1}{x}+c$
• C. $x\tan \frac{1}{x}+c$
• D. $\tan \frac{1}{x}+c$

Answer 1

The correct option is A $x^3\tan \frac{1}{x}+c$

$\int (3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$
$\int \left(3x^2\tan \frac{1}{x}\right)dx-\int x{\sec }^2\frac{1}{x}dx$
$\tan \frac{1}{x}\int 3x^{2}dx-\int {\sec }^2\frac{1}{x}\left(\frac{-1}{x^2}\right)x^{3}dx$
$\int x{\sec }^2\frac{1}{x}dx$
$=\tan \frac{1}{x}{x}^3+\int se{c}^2\frac{1}{x}-xdx$
$-\int xse{c}^2\frac{1}{x}dx+c$
$=\left(\tan \frac{1}{x}\right)x^3+c$
$\int (3x^2\tan \frac{1}{x}-xse{c}^2\frac{1}{x})dx=x^3\left(\tan \frac{1}{x}\right)+c$