Integral of $$$\frac{\left(\sqrt{x} - 1\right)^{2}}{\sqrt{x}}$$$

Find $$$\int \frac{\left(\sqrt{x} - 1\right)^{2}}{\sqrt{x}}\, dx$$$.

Let $$$u=\sqrt{x} - 1$$$.

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

Therefore,

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

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

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

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

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

Recall that $$$u=\sqrt{x} - 1$$$:

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

Therefore,

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

Add the constant of integration:

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

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