Tìm nguyên hàm √x dx (x > 0)
Giải thích
\(\int {\sqrt x } \;{\rm{d}}x = \int {{x^{\frac{1}{2}}}} \;{\rm{d}}x = \frac{{{x^{\frac{3}{2}}}}}{3} = \frac{{2x\sqrt x }}{3} + C\)
\(\int {\sqrt x } \;{\rm{d}}x = \int {{x^{\frac{1}{2}}}} \;{\rm{d}}x = \frac{{{x^{\frac{3}{2}}}}}{3} = \frac{{2x\sqrt x }}{3} + C\)