Tìm nguyên hàm: √x (7x^2 - 3) dx (x > 0)

Phån tích tương tưu, ta có:

\(\int {\sqrt x } \left( {7{x^2} - 3} \right){\rm{d}}x = \int 7 {x^{\frac{5}{2}}}\;{\rm{d}}x - \int 3 {x^{\frac{1}{2}}}\;{\rm{d}}x = 7 \cdot \frac{2}{7}{x^{\frac{7}{2}}} - 3 \cdot \frac{2}{3}{x^{\frac{3}{2}}} + C{\rm{. }}\)

Đap só: \(\int {\sqrt x } \left( {7{x^2} - 3} \right){\rm{d}}x = 2\left( {{x^3} - x} \right)\sqrt x  + C\).