# Thread: Radical-integral

1. ## Radical-integral

Calculate :

2. By rationalizing: $\frac{x}{x + \sqrt{x^2 - 1}} \cdot {\color{red} \frac{x-\sqrt{x^2 - 1}}{x - \sqrt{x^2-1}}} = \frac{x^2 - x\sqrt{x^2-1}}{x^2 - x^2 + 1} = x^2 - x\sqrt{x^2-1}$

Therefore, your problem is equivalent to: $2\int \left(x^2 - x\sqrt{x^2-1}\right)dx$

The first term is simple. The second one can be done with a u-sub: $u = x^2 - 1 \ \Rightarrow \ du = 2x dx$