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