Question

неопределенный интеграл​

Answer 1

$\int \dfrac{\sqrt{1+\sqrt{x}}}{\sqrt{x}}\, dx=\Big[\ t=\sqrt{x}\ ,\ x=t^2\ ,\ dx=2t\, dt\ \Big]=\int \dfrac{\sqrt{1+t}}{t}\cdot 2t\, dt=\\\\\\=2\int \sqrt{1+t}\, dt=2\cdot \dfrac{(1+t)^{3/2}}{3/2}+C=\dfrac{4}{3}\cdot \sqrt{(1+t)^3}+C=\\\\\\=\dfrac{4}{3}\cdot \sqrt{(1+\sqrt{x})^3}+C$