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 $x^3\tan \frac{1}{x}+c$

$I=\int (3x^2\tan \frac{1}{x}-x{\sec }^2\frac{1}{x})dx$

$I=\int 3x^2\tan \frac{1}{x}dx-\int {\sec }^2\frac{1}{x}.xdx$

$I=I_1-I_2$

$I_1=\int 3x^2\tan \frac{1}{x}dx$

$=\tan \frac{1}{x}.\int 3x^{2}dx-\int (\frac{d}{dx}\tan \frac{1}{x}.\int 3x^{2}dx)dx+c$

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

$=\tan \frac{1}{x}.x^3+\int x{\sec }^2\frac{1}{x}dx+c$

Now $I=I_1-I_2$

$=\tan \frac{1}{x}.x^3+\int x.{\sec }^2\frac{1}{x}dx-\int x{\sec }^2\frac{1}{x}dx+c$

$=x^2\tan \frac{1}{x}+c$.