Question

⁵√(3-2x) найти интеграл

Answer 1

$\int\sqrt[5]{3-2x}dx=\int(3-2x)^\frac15dx=-\frac12\int(3-2x)^\frac15d(3-2x)=$

$-\frac12\frac{(3-2x)^{{\displaystyle\frac15}+1}}{{\displaystyle\frac15}+1}+C=-\frac{5}{12}\left(3-2x\right)^{\frac{6}{5}}+C$