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

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

Rewrite the numerator as $$$x - 2=\left(x - 1\right) - 1$$$ and split the fraction:

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

Integrate term by term:

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

Let $$$u=x - 1$$$.

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

Thus,

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

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

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

Recall that $$$u=x - 1$$$:

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

Let $$$u=x - 1$$$.

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

Thus,

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

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

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

Recall that $$$u=x - 1$$$:

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

Therefore,

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

Simplify:

$$\int{\frac{x - 2}{\sqrt{x - 1}} d x} = \frac{2 \left(x - 4\right) \sqrt{x - 1}}{3}$$

Add the constant of integration:

$$\int{\frac{x - 2}{\sqrt{x - 1}} d x} = \frac{2 \left(x - 4\right) \sqrt{x - 1}}{3}+C$$

$$$\int \frac{x - 2}{\sqrt{x - 1}}\, dx = \frac{2 \left(x - 4\right) \sqrt{x - 1}}{3} + C$$$A