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Differentiate w.r.t. x
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\int 3x^{5}+15x^{4}+21x^{2}+18x^{3}+42x\mathrm{d}x
Use the distributive property to multiply 3x^{2}+6x by x^{3}+3x^{2}+7 and combine like terms.
\int 3x^{5}\mathrm{d}x+\int 15x^{4}\mathrm{d}x+\int 21x^{2}\mathrm{d}x+\int 18x^{3}\mathrm{d}x+\int 42x\mathrm{d}x
Integrate the sum term by term.
3\int x^{5}\mathrm{d}x+15\int x^{4}\mathrm{d}x+21\int x^{2}\mathrm{d}x+18\int x^{3}\mathrm{d}x+42\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{6}}{2}+15\int x^{4}\mathrm{d}x+21\int x^{2}\mathrm{d}x+18\int x^{3}\mathrm{d}x+42\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}. Multiply 3 times \frac{x^{6}}{6}.
\frac{x^{6}}{2}+3x^{5}+21\int x^{2}\mathrm{d}x+18\int x^{3}\mathrm{d}x+42\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^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply 15 times \frac{x^{5}}{5}.
\frac{x^{6}}{2}+3x^{5}+7x^{3}+18\int x^{3}\mathrm{d}x+42\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^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 21 times \frac{x^{3}}{3}.
\frac{x^{6}}{2}+3x^{5}+7x^{3}+\frac{9x^{4}}{2}+42\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}. Multiply 18 times \frac{x^{4}}{4}.
\frac{x^{6}}{2}+3x^{5}+7x^{3}+\frac{9x^{4}}{2}+21x^{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}. Multiply 42 times \frac{x^{2}}{2}.
\frac{x^{6}}{2}+3x^{5}+7x^{3}+\frac{9x^{4}}{2}+21x^{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.