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\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x+\int -20x\mathrm{d}x
Integrate the sum term by term.
\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{6}}{6}+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.
\frac{x^{6}}{6}+\frac{x^{4}}{4}-20\int x\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}.
\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}
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}. اضرب -20 في \frac{x^{2}}{2}.
\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}+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.