Evaluate
-\frac{2x^{2}}{5}+\frac{17x}{5}+С
Differentiate w.r.t. x
\frac{17-4x}{5}
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\int -\frac{4x}{5}\mathrm{d}x+\int \frac{17}{5}\mathrm{d}x
Integrate the sum term by term.
-\frac{4\int x\mathrm{d}x}{5}+\int \frac{17}{5}\mathrm{d}x
Factor out the constant in each of the terms.
-\frac{2x^{2}}{5}+\int \frac{17}{5}\mathrm{d}x
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 -\frac{4}{5} times \frac{x^{2}}{2}.
\frac{-2x^{2}+17x}{5}
Find the integral of \frac{17}{5} using the table of common integrals rule \int a\mathrm{d}x=ax.
-\frac{2x^{2}}{5}+\frac{17x}{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.
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