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\int \frac{1}{x^{2}}-\frac{1}{x^{3}}\mathrm{d}x
Evaluate the indefinite integral first.
\int \frac{1}{x^{2}}\mathrm{d}x+\int -\frac{1}{x^{3}}\mathrm{d}x
Integrate the sum term by term.
\int \frac{1}{x^{2}}\mathrm{d}x-\int \frac{1}{x^{3}}\mathrm{d}x
Factor out the constant in each of the terms.
-\frac{1}{x}-\int \frac{1}{x^{3}}\mathrm{d}x
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{2}}\mathrm{d}x with -\frac{1}{x}.
-\frac{1}{x}+\frac{1}{2x^{2}}
Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int \frac{1}{x^{3}}\mathrm{d}x with -\frac{1}{2x^{2}}. Multiply -1 times -\frac{1}{2x^{2}}.
\frac{\frac{1}{2}-x}{x^{2}}
Simplify.
\left(\frac{1}{2}-\left(-1\right)\right)\left(-1\right)^{-2}-\left(\frac{1}{2}-\left(-3\right)\right)\left(-3\right)^{-2}
The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.
\frac{10}{9}
Simplify.