Question

The value of integratin $\int x{\sec }^{-1}xdx$ is

• A. $\frac{1}{2}{x}^2{\sec }^{-1}x+\frac{1}{2}\sqrt{x^2-1}+c$
• B. $\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{1}{2}\sqrt{x^2-1}+c$
• C. $x^2{\sec }^{2}x+\frac{1}{2}\sqrt{x^2-1}+c$
• D. $x^2{\sec }^{2}x-\frac{1}{2}\sqrt{x^2-1}+c$

Answer 1

The correct option is B $\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{1}{2}\sqrt{x^2-1}+c$

The value of $\int x{\sec }^{-1}xdx$ is

$={\sec }^{-1}x\cdot \frac{x^2}{2}-\int \frac{1}{\sqrt{x^2-1}}\cdot \frac{x^2}{2}dx$

$=\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{1}{4}\int \frac{d{x}^2}{\sqrt{x^2-1}}dx$

$=\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{1}{2}\sqrt{x^2-1}+c$