Question

$\int x{\sec }^{-1}xdx=\frac{2}{k}[x^2{\sec }^{-1}x-\sqrt{x^2-1}].$ Find the value of $k$.

Answer 1

Let $I=\int x{\sec }^{-1}xdx$
$=\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{1}{2}\int \frac{x}{\sqrt{x^2-1}}dx$
Put $t=x^2-1\Rightarrow dt=2xdx$
$I=\frac{1}{2}{x}^2{\sec }^{-1}x\frac{-1}{4}\int \frac{1}{\sqrt{u}}du$
${\sec }^{-1}x-\frac{\sqrt{u}}{2}=\frac{1}{2}{x}^2{\sec }^{-1}x-\frac{\sqrt{x^2-1}}{2}$