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Differentiate w.r.t. x
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\int \left(x^{3}-3x^{2}+3x-1\right)\left(x-2\right)\mathrm{d}x
Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-1\right)^{3}.
\int x^{4}-5x^{3}+9x^{2}-7x+2\mathrm{d}x
Use the distributive property to multiply x^{3}-3x^{2}+3x-1 by x-2 and combine like terms.
\int x^{4}\mathrm{d}x+\int -5x^{3}\mathrm{d}x+\int 9x^{2}\mathrm{d}x+\int -7x\mathrm{d}x+\int 2\mathrm{d}x
Integrate the sum term by term.
\int x^{4}\mathrm{d}x-5\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-7\int x\mathrm{d}x+\int 2\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{5}}{5}-5\int x^{3}\mathrm{d}x+9\int x^{2}\mathrm{d}x-7\int x\mathrm{d}x+\int 2\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}.
\frac{x^{5}}{5}-\frac{5x^{4}}{4}+9\int x^{2}\mathrm{d}x-7\int x\mathrm{d}x+\int 2\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}. Multiply -5 times \frac{x^{4}}{4}.
\frac{x^{5}}{5}-\frac{5x^{4}}{4}+3x^{3}-7\int x\mathrm{d}x+\int 2\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 9 times \frac{x^{3}}{3}.
\frac{x^{5}}{5}-\frac{5x^{4}}{4}+3x^{3}-\frac{7x^{2}}{2}+\int 2\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -7 times \frac{x^{2}}{2}.
\frac{x^{5}}{5}-\frac{5x^{4}}{4}+3x^{3}-\frac{7x^{2}}{2}+2x
Find the integral of 2 using the table of common integrals rule \int a\mathrm{d}x=ax.
-\frac{7x^{2}}{2}+2x+3x^{3}-\frac{5x^{4}}{4}+\frac{x^{5}}{5}
Simplify.
-\frac{7x^{2}}{2}+2x+3x^{3}-\frac{5x^{4}}{4}+\frac{x^{5}}{5}+С
If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.