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

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

Let $$$u=2 x - 1$$$.

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

Thus,

$${\color{red}{\int{\frac{x}{\sqrt{2 x - 1}} d x}}} = {\color{red}{\int{\frac{u + 1}{4 \sqrt{u}} d u}}}$$

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

$${\color{red}{\int{\frac{u + 1}{4 \sqrt{u}} d u}}} = {\color{red}{\left(\frac{\int{\frac{u + 1}{\sqrt{u}} d u}}{4}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\frac{u + 1}{\sqrt{u}} d u}}}}{4} = \frac{{\color{red}{\int{\left(\sqrt{u} + \frac{1}{\sqrt{u}}\right)d u}}}}{4}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\sqrt{u} + \frac{1}{\sqrt{u}}\right)d u}}}}{4} = \frac{{\color{red}{\left(\int{\frac{1}{\sqrt{u}} d u} + \int{\sqrt{u} d u}\right)}}}{4}$$

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

$$\frac{\int{\frac{1}{\sqrt{u}} d u}}{4} + \frac{{\color{red}{\int{\sqrt{u} d u}}}}{4}=\frac{\int{\frac{1}{\sqrt{u}} d u}}{4} + \frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{4}=\frac{\int{\frac{1}{\sqrt{u}} d u}}{4} + \frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{4}=\frac{\int{\frac{1}{\sqrt{u}} d u}}{4} + \frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{4}$$

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

$$\frac{u^{\frac{3}{2}}}{6} + \frac{{\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{4}=\frac{u^{\frac{3}{2}}}{6} + \frac{{\color{red}{\int{u^{- \frac{1}{2}} d u}}}}{4}=\frac{u^{\frac{3}{2}}}{6} + \frac{{\color{red}{\frac{u^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}}{4}=\frac{u^{\frac{3}{2}}}{6} + \frac{{\color{red}{\left(2 u^{\frac{1}{2}}\right)}}}{4}=\frac{u^{\frac{3}{2}}}{6} + \frac{{\color{red}{\left(2 \sqrt{u}\right)}}}{4}$$

Recall that $$$u=2 x - 1$$$:

$$\frac{\sqrt{{\color{red}{u}}}}{2} + \frac{{\color{red}{u}}^{\frac{3}{2}}}{6} = \frac{\sqrt{{\color{red}{\left(2 x - 1\right)}}}}{2} + \frac{{\color{red}{\left(2 x - 1\right)}}^{\frac{3}{2}}}{6}$$

Therefore,

$$\int{\frac{x}{\sqrt{2 x - 1}} d x} = \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \frac{\sqrt{2 x - 1}}{2}$$

Simplify:

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

Add the constant of integration:

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

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