Question

Vreau să ştiu dacă este corectă rezolvarea :

$\int \: \frac{ x \sqrt[3]{x} + 2 {x}^{2} \sqrt[4]{ {x}^{2} } }{ \sqrt{x} } \: dx \: ,x > 0$

$= \int \frac{x \sqrt[3]{x} + 2 {x}^{2} \sqrt[4]{ {x}^{2} } }{ {x}^{ \frac{1}{2} } } = \int \frac{x \sqrt[3]{x} }{ {x}^{ \frac{1}{2} } } \: dx + \int \frac{2 {x}^{2} \sqrt[4]{ {x}^{2} } }{ {x}^{ \frac{1}{2} } }$

$= \int \frac{x \times {x}^{ \frac{1}{3} } }{ {x}^{ \frac{1}{2} } } \: dx + 2 \int \frac{ {x}^{2} \times {x}^{ \frac{2}{4} } }{ {x}^{ \frac{1}{2} } } \: dx$

$= \int \frac{ {x}^{ \frac{4}{3} } }{ {x}^{ \frac{1}{2} } } \: dx + 2 \int \frac{ {x}^{2} \times {x}^{ \frac{1}{2} } }{ {x}^{ \frac{1}{2} } } \: dx$

$= \int {x}^{ \frac{5}{6} } \: dx + 2 \times \frac{ {x}^{2 + 1} }{2 + 1}$

$= \frac{ {x}^{ \frac{5}{6} + 1} }{ \frac{5}{6} + 1} + 2 \times \frac{ {x}^{3} }{3}$

$= \frac{ {x}^{ \frac{11}{6} } }{ \frac{11}{6} } + \frac{2 {x}^{3} }{3} + C$

Answer 1

corectă rezolvarea.......