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Differentiate w.r.t. x
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\int \left(2x-x^{2}\right)^{2}\mathrm{d}x
Combine 4x and -2x to get 2x.
\int 4x^{2}-4xx^{2}+\left(x^{2}\right)^{2}\mathrm{d}x
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-x^{2}\right)^{2}.
\int 4x^{2}-4x^{3}+\left(x^{2}\right)^{2}\mathrm{d}x
To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.
\int 4x^{2}-4x^{3}+x^{4}\mathrm{d}x
To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.
\int 4x^{2}\mathrm{d}x+\int -4x^{3}\mathrm{d}x+\int x^{4}\mathrm{d}x
Integrate the sum term by term.
4\int x^{2}\mathrm{d}x-4\int x^{3}\mathrm{d}x+\int x^{4}\mathrm{d}x
Factor out the constant in each of the terms.
\frac{4x^{3}}{3}-4\int x^{3}\mathrm{d}x+\int x^{4}\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 4 times \frac{x^{3}}{3}.
\frac{4x^{3}}{3}-x^{4}+\int x^{4}\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 -4 times \frac{x^{4}}{4}.
\frac{4x^{3}}{3}-x^{4}+\frac{x^{5}}{5}
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}-x^{4}+\frac{4x^{3}}{3}
Simplify.
\frac{x^{5}}{5}-x^{4}+\frac{4x^{3}}{3}+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.