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

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

Let $$$u=\sqrt{x}$$$.

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \frac{u^{2}}{1 - u}$$$:

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

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Let $$$v=1 - u$$$.

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

Thus,

$$- 2 u - 2 \int{u d u} + 2 {\color{red}{\int{\frac{1}{1 - u} d u}}} = - 2 u - 2 \int{u d u} + 2 {\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$- 2 u - 2 \int{u d u} + 2 {\color{red}{\int{\left(- \frac{1}{v}\right)d v}}} = - 2 u - 2 \int{u d u} + 2 {\color{red}{\left(- \int{\frac{1}{v} d v}\right)}}$$

The integral of $$$\frac{1}{v}$$$ is $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

$$- 2 u - 2 \int{u d u} - 2 {\color{red}{\int{\frac{1}{v} d v}}} = - 2 u - 2 \int{u d u} - 2 {\color{red}{\ln{\left(\left|{v}\right| \right)}}}$$

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

$$- 2 u - 2 \ln{\left(\left|{{\color{red}{v}}}\right| \right)} - 2 \int{u d u} = - 2 u - 2 \ln{\left(\left|{{\color{red}{\left(1 - u\right)}}}\right| \right)} - 2 \int{u d u}$$

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

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

Recall that $$$u=\sqrt{x}$$$:

$$- 2 \ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)} - 2 {\color{red}{u}} - {\color{red}{u}}^{2} = - 2 \ln{\left(\left|{-1 + {\color{red}{\sqrt{x}}}}\right| \right)} - 2 {\color{red}{\sqrt{x}}} - {\color{red}{\sqrt{x}}}^{2}$$

Therefore,

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

Add the constant of integration:

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

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