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تفاضل w.r.t. x
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2\int x^{22}\mathrm{d}x
Factor out the constant using \int af\left(x\right)\mathrm{d}x=a\int f\left(x\right)\mathrm{d}x.
\frac{2x^{23}}{23}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{22}\mathrm{d}x with \frac{x^{23}}{23}.
\frac{2x^{23}}{23}+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.