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

Find $$$\int \frac{y^{3}}{1 - y}\, dy$$$.

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

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dy = c y$$$ with $$$c=1$$$:

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

Let $$$u=1 - y$$$.

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

The integral becomes

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

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

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

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

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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