Integral of $$$\frac{\sqrt{x - 1}}{x}$$$

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

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

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

So,

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

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

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

The integral of $$$\frac{1}{u^{2} + 1}$$$ is $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

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

$$- 2 \operatorname{atan}{\left({\color{red}{u}} \right)} + 2 {\color{red}{u}} = - 2 \operatorname{atan}{\left({\color{red}{\sqrt{x - 1}}} \right)} + 2 {\color{red}{\sqrt{x - 1}}}$$

Therefore,

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

Add the constant of integration:

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

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