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Differentiate w.r.t. x
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\int \frac{x\left(x-2\right)\left(x+2\right)\left(x^{2}+5\right)}{x+2}\mathrm{d}x
Factor the expressions that are not already factored in \frac{x^{5}+x^{3}-20x}{x+2}.
\int x\left(x-2\right)\left(x^{2}+5\right)\mathrm{d}x
Cancel out x+2 in both numerator and denominator.
\int x^{4}-2x^{3}+5x^{2}-10x\mathrm{d}x
Expand the expression.
\int x^{4}\mathrm{d}x+\int -2x^{3}\mathrm{d}x+\int 5x^{2}\mathrm{d}x+\int -10x\mathrm{d}x
Integrate the sum term by term.
\int x^{4}\mathrm{d}x-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\int x\mathrm{d}x
Factor out the constant in each of the terms.
\frac{x^{5}}{5}-2\int x^{3}\mathrm{d}x+5\int x^{2}\mathrm{d}x-10\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}.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+5\int x^{2}\mathrm{d}x-10\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 -2 times \frac{x^{4}}{4}.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-10\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 5 times \frac{x^{3}}{3}.
\frac{x^{5}}{5}-\frac{x^{4}}{2}+\frac{5x^{3}}{3}-5x^{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 -10 times \frac{x^{2}}{2}.
-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\frac{x^{5}}{5}
Simplify.
-5x^{2}+\frac{5x^{3}}{3}-\frac{x^{4}}{2}+\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.