Integral of $$$\frac{x}{\sqrt[4]{1 - x}}$$$

Find $$$\int \frac{x}{\sqrt[4]{1 - x}}\, dx$$$.

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

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

Therefore,

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

Expand the expression:

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

Integrate term by term:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{3}{4}$$$:

$$- \int{\frac{1}{\sqrt[4]{u}} d u} + {\color{red}{\int{u^{\frac{3}{4}} d u}}}=- \int{\frac{1}{\sqrt[4]{u}} d u} + {\color{red}{\frac{u^{\frac{3}{4} + 1}}{\frac{3}{4} + 1}}}=- \int{\frac{1}{\sqrt[4]{u}} d u} + {\color{red}{\left(\frac{4 u^{\frac{7}{4}}}{7}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{1}{4}$$$:

$$\frac{4 u^{\frac{7}{4}}}{7} - {\color{red}{\int{\frac{1}{\sqrt[4]{u}} d u}}}=\frac{4 u^{\frac{7}{4}}}{7} - {\color{red}{\int{u^{- \frac{1}{4}} d u}}}=\frac{4 u^{\frac{7}{4}}}{7} - {\color{red}{\frac{u^{- \frac{1}{4} + 1}}{- \frac{1}{4} + 1}}}=\frac{4 u^{\frac{7}{4}}}{7} - {\color{red}{\left(\frac{4 u^{\frac{3}{4}}}{3}\right)}}$$

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

$$- \frac{4 {\color{red}{u}}^{\frac{3}{4}}}{3} + \frac{4 {\color{red}{u}}^{\frac{7}{4}}}{7} = - \frac{4 {\color{red}{\left(1 - x\right)}}^{\frac{3}{4}}}{3} + \frac{4 {\color{red}{\left(1 - x\right)}}^{\frac{7}{4}}}{7}$$

Therefore,

$$\int{\frac{x}{\sqrt[4]{1 - x}} d x} = \frac{4 \left(1 - x\right)^{\frac{7}{4}}}{7} - \frac{4 \left(1 - x\right)^{\frac{3}{4}}}{3}$$

Simplify:

$$\int{\frac{x}{\sqrt[4]{1 - x}} d x} = \frac{4 \left(1 - x\right)^{\frac{3}{4}} \left(- 3 x - 4\right)}{21}$$

Add the constant of integration:

$$\int{\frac{x}{\sqrt[4]{1 - x}} d x} = \frac{4 \left(1 - x\right)^{\frac{3}{4}} \left(- 3 x - 4\right)}{21}+C$$

$$$\int \frac{x}{\sqrt[4]{1 - x}}\, dx = \frac{4 \left(1 - x\right)^{\frac{3}{4}} \left(- 3 x - 4\right)}{21} + C$$$A