Tìm nguyên hàm 1/x^3 dx
Giải thích
\(\int {\frac{1}{{{x^3}}}} \;{\rm{d}}x = \int {\left( {{x^{ - 3}}} \right)} {\rm{d}}x = \frac{{{x^{ - 2}}}}{{ - 2}} + C = - \frac{1}{{2{x^2}}} + C\)
\(\int {\frac{1}{{{x^3}}}} \;{\rm{d}}x = \int {\left( {{x^{ - 3}}} \right)} {\rm{d}}x = \frac{{{x^{ - 2}}}}{{ - 2}} + C = - \frac{1}{{2{x^2}}} + C\)