Integral of $$$\frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}}$$$

Find $$$\int \frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}}\, dx$$$.

Simplify the integrand:

$${\color{red}{\int{\frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}} d x}}} = {\color{red}{\int{\frac{1}{3 x^{4} \left(1 - x\right)^{2}} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{4} \left(1 - x\right)^{2}}$$$:

$${\color{red}{\int{\frac{1}{3 x^{4} \left(1 - x\right)^{2}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{4} \left(1 - x\right)^{2}} d x}}{3}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{x^{4} \left(1 - x\right)^{2}} d x}}}}{3} = \frac{{\color{red}{\int{\left(- \frac{4}{x - 1} + \frac{1}{\left(x - 1\right)^{2}} + \frac{4}{x} + \frac{3}{x^{2}} + \frac{2}{x^{3}} + \frac{1}{x^{4}}\right)d x}}}}{3}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(- \frac{4}{x - 1} + \frac{1}{\left(x - 1\right)^{2}} + \frac{4}{x} + \frac{3}{x^{2}} + \frac{2}{x^{3}} + \frac{1}{x^{4}}\right)d x}}}}{3} = \frac{{\color{red}{\left(\int{\frac{1}{x^{4}} d x} + \int{\frac{2}{x^{3}} d x} + \int{\frac{3}{x^{2}} d x} + \int{\frac{4}{x} d x} + \int{\frac{1}{\left(x - 1\right)^{2}} d x} - \int{\frac{4}{x - 1} d x}\right)}}}{3}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{x^{4}} d x}}}}{3}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{x^{-4} d x}}}}{3}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}}{3}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\left(- \frac{x^{-3}}{3}\right)}}}{3}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}}{3}$$

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

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

So,

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}}}{3} - \frac{1}{9 x^{3}} = \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{3} - \frac{1}{9 x^{3}}$$

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

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{3} - \frac{1}{9 x^{3}}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\int{u^{-2} d u}}}}{3} - \frac{1}{9 x^{3}}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{3} - \frac{1}{9 x^{3}}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\left(- u^{-1}\right)}}}{3} - \frac{1}{9 x^{3}}=\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} + \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{3} - \frac{1}{9 x^{3}}$$

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

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} - \frac{{\color{red}{u}}^{-1}}{3} - \frac{1}{9 x^{3}} = \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{\int{\frac{4}{x - 1} d x}}{3} - \frac{{\color{red}{\left(x - 1\right)}}^{-1}}{3} - \frac{1}{9 x^{3}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{{\color{red}{\int{\frac{4}{x - 1} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}} = \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{{\color{red}{\left(4 \int{\frac{1}{x - 1} d x}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

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 becomes

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{4 {\color{red}{\int{\frac{1}{x - 1} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}} = \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{4 {\color{red}{\int{\frac{1}{u} d u}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

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

$$\frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{4 {\color{red}{\int{\frac{1}{u} d u}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}} = \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{4 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

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

$$- \frac{4 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{3} + \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}} = - \frac{4 \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{3} + \frac{\int{\frac{2}{x^{3}} d x}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

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

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{{\color{red}{\int{\frac{2}{x^{3}} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}} = - \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{{\color{red}{\left(2 \int{\frac{1}{x^{3}} d x}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{2 {\color{red}{\int{\frac{1}{x^{3}} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{2 {\color{red}{\int{x^{-3} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{2 {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{2 {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{3}{x^{2}} d x}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{2 {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{9 x^{3}}$$

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

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{{\color{red}{\int{\frac{3}{x^{2}} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}} = - \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + \frac{{\color{red}{\left(3 \int{\frac{1}{x^{2}} d x}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}$$

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

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + {\color{red}{\int{\frac{1}{x^{2}} d x}}} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + {\color{red}{\int{x^{-2} d x}}} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + {\color{red}{\left(- x^{-1}\right)}} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}=- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{\int{\frac{4}{x} d x}}{3} + {\color{red}{\left(- \frac{1}{x}\right)}} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{{\color{red}{\int{\frac{4}{x} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{x} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}} = - \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{{\color{red}{\left(4 \int{\frac{1}{x} d x}\right)}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{x} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}$$

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

$$- \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{4 {\color{red}{\int{\frac{1}{x} d x}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{x} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}} = - \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} + \frac{4 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{x} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}$$

Therefore,

$$\int{\frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}} d x} = \frac{4 \ln{\left(\left|{x}\right| \right)}}{3} - \frac{4 \ln{\left(\left|{x - 1}\right| \right)}}{3} - \frac{1}{3 \left(x - 1\right)} - \frac{1}{x} - \frac{1}{3 x^{2}} - \frac{1}{9 x^{3}}$$

Simplify:

$$\int{\frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}} d x} = \frac{12 x^{3} \left(x - 1\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 1}\right| \right)}\right) - 3 x^{3} + 9 x^{2} \left(1 - x\right) + 3 x \left(1 - x\right) - x + 1}{9 x^{3} \left(x - 1\right)}$$

Add the constant of integration:

$$\int{\frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}} d x} = \frac{12 x^{3} \left(x - 1\right) \left(\ln{\left(\left|{x}\right| \right)} - \ln{\left(\left|{x - 1}\right| \right)}\right) - 3 x^{3} + 9 x^{2} \left(1 - x\right) + 3 x \left(1 - x\right) - x + 1}{9 x^{3} \left(x - 1\right)}+C$$

$$$\int \frac{1}{3 \left(- x^{3} + x^{2}\right)^{2}}\, dx = \frac{12 x^{3} \left(x - 1\right) \left(\ln\left(\left|{x}\right|\right) - \ln\left(\left|{x - 1}\right|\right)\right) - 3 x^{3} + 9 x^{2} \left(1 - x\right) + 3 x \left(1 - x\right) - x + 1}{9 x^{3} \left(x - 1\right)} + C$$$A