Integral of $$$f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right)$$$ with respect to $$$x$$$

Find $$$\int f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=f^{2} int_{x}$$$ and $$$f{\left(x \right)} = int_{0}^{x} x^{2} \left(1 - x\right)$$$:

$${\color{red}{\int{f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right) d x}}} = {\color{red}{f^{2} int_{x} \int{int_{0}^{x} x^{2} \left(1 - x\right) d x}}}$$

For the integral $$$\int{int_{0}^{x} x^{2} \left(1 - x\right) d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x^{2} \left(1 - x\right)$$$ and $$$\operatorname{dv}=int_{0}^{x} dx$$$.

Then $$$\operatorname{du}=\left(x^{2} \left(1 - x\right)\right)^{\prime }dx=x \left(2 - 3 x\right) dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{int_{0}^{x} d x}=\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}$$$ (steps can be seen »).

So,

$$f^{2} int_{x} {\color{red}{\int{int_{0}^{x} x^{2} \left(1 - x\right) d x}}}=f^{2} int_{x} {\color{red}{\left(x^{2} \left(1 - x\right) \cdot \frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}-\int{\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}} \cdot x \left(2 - 3 x\right) d x}\right)}}=f^{2} int_{x} {\color{red}{\left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \int{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{\ln{\left(int_{0} \right)}}$$$ and $$$f{\left(x \right)} = int_{0}^{x} x \left(2 - 3 x\right)$$$:

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\int{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} d x}}}\right) = f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\frac{\int{int_{0}^{x} x \left(2 - 3 x\right) d x}}{\ln{\left(int_{0} \right)}}}}\right)$$

For the integral $$$\int{int_{0}^{x} x \left(2 - 3 x\right) d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x \left(2 - 3 x\right)$$$ and $$$\operatorname{dv}=int_{0}^{x} dx$$$.

Then $$$\operatorname{du}=\left(x \left(2 - 3 x\right)\right)^{\prime }dx=\left(2 - 6 x\right) dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{int_{0}^{x} d x}=\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}$$$ (steps can be seen »).

Therefore,

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{{\color{red}{\int{int_{0}^{x} x \left(2 - 3 x\right) d x}}}}{\ln{\left(int_{0} \right)}}\right)=f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{{\color{red}{\left(x \left(2 - 3 x\right) \cdot \frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}-\int{\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}} \cdot \left(2 - 6 x\right) d x}\right)}}}{\ln{\left(int_{0} \right)}}\right)=f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{{\color{red}{\left(\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} - \int{\left(- \frac{2 int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}}\right)d x}\right)}}}{\ln{\left(int_{0} \right)}}\right)$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=- \frac{2}{\ln{\left(int_{0} \right)}}$$$ and $$$f{\left(x \right)} = int_{0}^{x} \left(3 x - 1\right)$$$:

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\int{\left(- \frac{2 int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}}\right)d x}}}}{\ln{\left(int_{0} \right)}}\right) = f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\left(- \frac{2 \int{int_{0}^{x} \left(3 x - 1\right) d x}}{\ln{\left(int_{0} \right)}}\right)}}}{\ln{\left(int_{0} \right)}}\right)$$

For the integral $$$\int{int_{0}^{x} \left(3 x - 1\right) d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=3 x - 1$$$ and $$$\operatorname{dv}=int_{0}^{x} dx$$$.

Then $$$\operatorname{du}=\left(3 x - 1\right)^{\prime }dx=3 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{int_{0}^{x} d x}=\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}$$$ (steps can be seen »).

The integral can be rewritten as

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 {\color{red}{\int{int_{0}^{x} \left(3 x - 1\right) d x}}}}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)=f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 {\color{red}{\left(\left(3 x - 1\right) \cdot \frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}-\int{\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}} \cdot 3 d x}\right)}}}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)=f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 {\color{red}{\left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - \int{\frac{3 int_{0}^{x}}{\ln{\left(int_{0} \right)}} d x}\right)}}}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{3}{\ln{\left(int_{0} \right)}}$$$ and $$$f{\left(x \right)} = int_{0}^{x}$$$:

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 \left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\int{\frac{3 int_{0}^{x}}{\ln{\left(int_{0} \right)}} d x}}}\right)}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right) = f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 \left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - {\color{red}{\left(\frac{3 \int{int_{0}^{x} d x}}{\ln{\left(int_{0} \right)}}\right)}}\right)}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=int_{0}$$$:

$$f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 \left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - \frac{3 {\color{red}{\int{int_{0}^{x} d x}}}}{\ln{\left(int_{0} \right)}}\right)}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right) = f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 \left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - \frac{3 {\color{red}{\frac{int_{0}^{x}}{\ln{\left(int_{0} \right)}}}}}{\ln{\left(int_{0} \right)}}\right)}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)$$

Therefore,

$$\int{f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right) d x} = f^{2} int_{x} \left(\frac{int_{0}^{x} x^{2} \left(1 - x\right)}{\ln{\left(int_{0} \right)}} - \frac{\frac{int_{0}^{x} x \left(2 - 3 x\right)}{\ln{\left(int_{0} \right)}} + \frac{2 \left(\frac{int_{0}^{x} \left(3 x - 1\right)}{\ln{\left(int_{0} \right)}} - \frac{3 int_{0}^{x}}{\ln{\left(int_{0} \right)}^{2}}\right)}{\ln{\left(int_{0} \right)}}}{\ln{\left(int_{0} \right)}}\right)$$

Simplify:

$$\int{f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right) d x} = - \frac{f^{2} int_{0}^{x} int_{x} \left(x^{2} \left(x - 1\right) \ln{\left(int_{0} \right)}^{3} - x \left(3 x - 2\right) \ln{\left(int_{0} \right)}^{2} + 2 \left(3 x - 1\right) \ln{\left(int_{0} \right)} - 6\right)}{\ln{\left(int_{0} \right)}^{4}}$$

Add the constant of integration:

$$\int{f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right) d x} = - \frac{f^{2} int_{0}^{x} int_{x} \left(x^{2} \left(x - 1\right) \ln{\left(int_{0} \right)}^{3} - x \left(3 x - 2\right) \ln{\left(int_{0} \right)}^{2} + 2 \left(3 x - 1\right) \ln{\left(int_{0} \right)} - 6\right)}{\ln{\left(int_{0} \right)}^{4}}+C$$

$$$\int f^{2} int_{0}^{x} int_{x} x^{2} \left(1 - x\right)\, dx = - \frac{f^{2} int_{0}^{x} int_{x} \left(x^{2} \left(x - 1\right) \ln^{3}\left(int_{0}\right) - x \left(3 x - 2\right) \ln^{2}\left(int_{0}\right) + 2 \left(3 x - 1\right) \ln\left(int_{0}\right) - 6\right)}{\ln^{4}\left(int_{0}\right)} + C$$$A