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

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

Let $$$u=4 - x$$$.

Then $$$du=\left(4 - x\right)^{\prime }dx = - dx$$$ (steps can be seen »), and we have that $$$dx = - du$$$.

So,

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

Expand the expression:

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

Integrate term by term:

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

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

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

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

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

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

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

Recall that $$$u=4 - x$$$:

$$- 8 \sqrt{{\color{red}{u}}} + \frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = - 8 \sqrt{{\color{red}{\left(4 - x\right)}}} + \frac{2 {\color{red}{\left(4 - x\right)}}^{\frac{3}{2}}}{3}$$

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \frac{x}{\sqrt{4 - x}}\, dx = \frac{2 \sqrt{4 - x} \left(- x - 8\right)}{3} + C$$$A