Question

The value of the integral $\int (3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx,$ 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. none of these

Answer 1

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

Let $I=\int (3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$
$\Rightarrow I=\int {\underset{II}{3x}}^2\underset{I}{\tan }\frac{1}{x}dx-\int x{\sec }^2\frac{1}{x}dx$
$\Rightarrow I=x^3\tan \frac{1}{x}+\int x{\sec }^2\frac{1}{x}dx-\int x{\sec }^2\frac{1}{x}dx+C$
$=x^3\tan \frac{1}{x}+C$