Integral of $$$\frac{x}{\sqrt{4 x - 6}}$$$

Find $$$\int \frac{x}{\sqrt{4 x - 6}}\, dx$$$.

Simplify the integrand:

$${\color{red}{\int{\frac{x}{\sqrt{4 x - 6}} d x}}} = {\color{red}{\int{\frac{\sqrt{2} x}{2 \sqrt{2 x - 3}} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{\sqrt{2}}{2}$$$ and $$$f{\left(x \right)} = \frac{x}{\sqrt{2 x - 3}}$$$:

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

Let $$$u=2 x - 3$$$.

Then $$$du=\left(2 x - 3\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

The integral can be rewritten as

$$\frac{\sqrt{2} {\color{red}{\int{\frac{x}{\sqrt{2 x - 3}} d x}}}}{2} = \frac{\sqrt{2} {\color{red}{\int{\frac{u + 3}{4 \sqrt{u}} d u}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \frac{u + 3}{\sqrt{u}}$$$:

$$\frac{\sqrt{2} {\color{red}{\int{\frac{u + 3}{4 \sqrt{u}} d u}}}}{2} = \frac{\sqrt{2} {\color{red}{\left(\frac{\int{\frac{u + 3}{\sqrt{u}} d u}}{4}\right)}}}{2}$$

Expand the expression:

$$\frac{\sqrt{2} {\color{red}{\int{\frac{u + 3}{\sqrt{u}} d u}}}}{8} = \frac{\sqrt{2} {\color{red}{\int{\left(\sqrt{u} + \frac{3}{\sqrt{u}}\right)d u}}}}{8}$$

Integrate term by term:

$$\frac{\sqrt{2} {\color{red}{\int{\left(\sqrt{u} + \frac{3}{\sqrt{u}}\right)d u}}}}{8} = \frac{\sqrt{2} {\color{red}{\left(\int{\frac{3}{\sqrt{u}} d u} + \int{\sqrt{u} d u}\right)}}}{8}$$

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

$$\frac{\sqrt{2} \left(\int{\frac{3}{\sqrt{u}} d u} + {\color{red}{\int{\sqrt{u} d u}}}\right)}{8}=\frac{\sqrt{2} \left(\int{\frac{3}{\sqrt{u}} d u} + {\color{red}{\int{u^{\frac{1}{2}} d u}}}\right)}{8}=\frac{\sqrt{2} \left(\int{\frac{3}{\sqrt{u}} d u} + {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}\right)}{8}=\frac{\sqrt{2} \left(\int{\frac{3}{\sqrt{u}} d u} + {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}\right)}{8}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=3$$$ and $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

$$\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + {\color{red}{\int{\frac{3}{\sqrt{u}} d u}}}\right)}{8} = \frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + {\color{red}{\left(3 \int{\frac{1}{\sqrt{u}} d u}\right)}}\right)}{8}$$

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

$$\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + 3 {\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}\right)}{8}=\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + 3 {\color{red}{\int{u^{- \frac{1}{2}} d u}}}\right)}{8}=\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + 3 {\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}\right)}{8}=\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + 3 {\color{red}{\left(2 u^{\frac{1}{2}}\right)}}\right)}{8}=\frac{\sqrt{2} \left(\frac{2 u^{\frac{3}{2}}}{3} + 3 {\color{red}{\left(2 \sqrt{u}\right)}}\right)}{8}$$

Recall that $$$u=2 x - 3$$$:

$$\frac{\sqrt{2} \left(6 \sqrt{{\color{red}{u}}} + \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3}\right)}{8} = \frac{\sqrt{2} \left(6 \sqrt{{\color{red}{\left(2 x - 3\right)}}} + \frac{2 {\color{red}{\left(2 x - 3\right)}}^{\frac{3}{2}}}{3}\right)}{8}$$

Therefore,

$$\int{\frac{x}{\sqrt{4 x - 6}} d x} = \frac{\sqrt{2} \left(\frac{2 \left(2 x - 3\right)^{\frac{3}{2}}}{3} + 6 \sqrt{2 x - 3}\right)}{8}$$

Simplify:

$$\int{\frac{x}{\sqrt{4 x - 6}} d x} = \left(\frac{x}{6} + \frac{1}{2}\right) \sqrt{4 x - 6}$$

Add the constant of integration:

$$\int{\frac{x}{\sqrt{4 x - 6}} d x} = \left(\frac{x}{6} + \frac{1}{2}\right) \sqrt{4 x - 6}+C$$

$$$\int \frac{x}{\sqrt{4 x - 6}}\, dx = \left(\frac{x}{6} + \frac{1}{2}\right) \sqrt{4 x - 6} + C$$$A