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

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

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,

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

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

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

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

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

Integrate term by term:

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

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

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

Perform partial fraction decomposition (steps can be seen »):

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

Integrate term by term:

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

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

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

Let $$$v=u + 1$$$.

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

So,

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

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

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

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

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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