Question

Найти неопр.интеграл
1) и 2)

Answer 1

$\int\limits {\frac{\sqrt{x}- \sqrt[3]{x^2} -x^{-\frac{1}{2}}}{x\sqrt{x}}} \, dx = \int\limits {\frac{x^{\frac{1}{2}}- x^{\frac{2}{3}} -x^{-\frac{1}{2}}}{x^{\frac{3}{2}}}} \, dx =\\\\ = \int\limits {[x^{\frac{1}{2}-\frac{3}{2}}- x^{\frac{2}{3}-\frac{3}{2}} -x^{-\frac{1}{2}-\frac{3}{2}}]} \, dx =\\\\ = \int\limits {[x^{-1}- x^{-\frac{5}{6}} -x^{-2}}]} \, dx =\\\\ = \int\limits {\frac{1}{x}} \, dx - \int\limits {x^{-\frac{5}{6}}} \, dx - \int\limits {x^{-2}} \, dx =\\\\ =ln|x|-6x^{\frac{1}{6}}+x^{-1}+C$

$\int\limits {\frac{x^5+3x^3}{x^2}} \, dx = \int\limits {[x^3+3x]} \, dx =\\\\ =\frac{x^4}{4}+\frac{3x^2}{2}+C$