Integral of $$$a^{2} f^{2} x^{2} + b^{2} f$$$ with respect to $$$x$$$
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Find $$$\int \left(a^{2} f^{2} x^{2} + b^{2} f\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(a^{2} f^{2} x^{2} + b^{2} f\right)d x}}} = {\color{red}{\left(\int{b^{2} f d x} + \int{a^{2} f^{2} x^{2} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=b^{2} f$$$:
$$\int{a^{2} f^{2} x^{2} d x} + {\color{red}{\int{b^{2} f d x}}} = \int{a^{2} f^{2} x^{2} d x} + {\color{red}{b^{2} f x}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=a^{2} f^{2}$$$ and $$$f{\left(x \right)} = x^{2}$$$:
$$b^{2} f x + {\color{red}{\int{a^{2} f^{2} x^{2} d x}}} = b^{2} f x + {\color{red}{a^{2} f^{2} \int{x^{2} d x}}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:
$$a^{2} f^{2} {\color{red}{\int{x^{2} d x}}} + b^{2} f x=a^{2} f^{2} {\color{red}{\frac{x^{1 + 2}}{1 + 2}}} + b^{2} f x=a^{2} f^{2} {\color{red}{\left(\frac{x^{3}}{3}\right)}} + b^{2} f x$$
Therefore,
$$\int{\left(a^{2} f^{2} x^{2} + b^{2} f\right)d x} = \frac{a^{2} f^{2} x^{3}}{3} + b^{2} f x$$
Simplify:
$$\int{\left(a^{2} f^{2} x^{2} + b^{2} f\right)d x} = \frac{f x \left(a^{2} f x^{2} + 3 b^{2}\right)}{3}$$
Add the constant of integration:
$$\int{\left(a^{2} f^{2} x^{2} + b^{2} f\right)d x} = \frac{f x \left(a^{2} f x^{2} + 3 b^{2}\right)}{3}+C$$
Answer
$$$\int \left(a^{2} f^{2} x^{2} + b^{2} f\right)\, dx = \frac{f x \left(a^{2} f x^{2} + 3 b^{2}\right)}{3} + C$$$A