Integral of $$$\frac{x^{3}}{x - 1}$$$

Find $$$\int \frac{x^{3}}{x - 1}\, dx$$$.

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

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

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

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

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

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

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

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

The integral can be rewritten as

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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