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

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

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 $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{x}{\sqrt{x} - 1} d x}}} = {\color{red}{\int{\frac{2 u^{3}}{u - 1} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \frac{u^{3}}{u - 1}$$$:

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

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

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

Integrate term by term:

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

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

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

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

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

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

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

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,

$$\frac{2 u^{3}}{3} + u^{2} + 2 u + 2 {\color{red}{\int{\frac{1}{u - 1} d u}}} = \frac{2 u^{3}}{3} + u^{2} + 2 u + 2 {\color{red}{\int{\frac{1}{v} d v}}}$$

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

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

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

$$\frac{2 u^{3}}{3} + u^{2} + 2 u + 2 \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = \frac{2 u^{3}}{3} + u^{2} + 2 u + 2 \ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}$$

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

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

Therefore,

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

Add the constant of integration:

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

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