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Expand the integral $\int\left(\frac{2}{x^5}+\frac{6}{x^2}+\frac{-3}{x}\right)dx$ into $3$ integrals using the sum rule for integrals, to then solve each integral separately
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$\int\frac{2}{x^5}dx+\int\frac{6}{x^2}dx+\int\frac{-3}{x}dx$
Learn how to solve problems step by step online. Integrate int(2/(x^5)+6/(x^2)-3/x)dx. Expand the integral \int\left(\frac{2}{x^5}+\frac{6}{x^2}+\frac{-3}{x}\right)dx into 3 integrals using the sum rule for integrals, to then solve each integral separately. The integral \int\frac{2}{x^5}dx results in: \frac{-1}{2x^{4}}. The integral \int\frac{6}{x^2}dx results in: \frac{-6}{x}. The integral \int\frac{-3}{x}dx results in: -3\ln\left(x\right).
