Integral of $$$x \sqrt{x^{2} - 1}$$$

Find $$$\int x \sqrt{x^{2} - 1}\, dx$$$.

Let $$$u=x^{2} - 1$$$.

Then $$$du=\left(x^{2} - 1\right)^{\prime }dx = 2 x dx$$$ (steps can be seen »), and we have that $$$x dx = \frac{du}{2}$$$.

The integral becomes

$${\color{red}{\int{x \sqrt{x^{2} - 1} d x}}} = {\color{red}{\int{\frac{\sqrt{u}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\frac{\sqrt{u}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sqrt{u} d u}}{2}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\frac{{\color{red}{\int{\sqrt{u} d u}}}}{2}=\frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{2}=\frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{2}=\frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{2}$$

Recall that $$$u=x^{2} - 1$$$:

$$\frac{{\color{red}{u}}^{\frac{3}{2}}}{3} = \frac{{\color{red}{\left(x^{2} - 1\right)}}^{\frac{3}{2}}}{3}$$

Therefore,

$$\int{x \sqrt{x^{2} - 1} d x} = \frac{\left(x^{2} - 1\right)^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{x \sqrt{x^{2} - 1} d x} = \frac{\left(x^{2} - 1\right)^{\frac{3}{2}}}{3}+C$$

$$$\int x \sqrt{x^{2} - 1}\, dx = \frac{\left(x^{2} - 1\right)^{\frac{3}{2}}}{3} + C$$$A