Integral of $$$- 4 x^{3} - \frac{x^{2}}{3} + x$$$

The calculator will find the integral/antiderivative of $$$- 4 x^{3} - \frac{x^{2}}{3} + x$$$, with steps shown.

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Find $$$\int \left(- 4 x^{3} - \frac{x^{2}}{3} + x\right)\, dx$$$.

Solution

Integrate term by term:

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

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

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

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

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

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

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

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 $$$f{\left(x \right)} = x^{2}$$$:

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

Answer

$$$\int \left(- 4 x^{3} - \frac{x^{2}}{3} + x\right)\, dx = x^{2} \left(- x^{2} - \frac{x}{9} + \frac{1}{2}\right) + C$$$A


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