Bài 2. Tìma) \(\int {\left( {\sqrt x + \root 3 \of x } \right)dx;} \) b) \(\int {{{x\sqrt x + \sqrt x } \over {{x^2}}}} dx;\)
c) \(\int {4{{\sin }^2}xdx;} \) d) \(\int {{{1 + \cos 4x} \over 2}dx.} \)Giảia) \(\int {\left( {\sqrt x + \root 3 \of x } \right)dx = \int {\left( {{x^{{1 \over 2}}} + {x^{{1 \over 3}}}} \right)dx = {{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} } + {{{x^{{4 \over 3}}}} \over {{4 \over 3}}} + C = {2 \over 3}{x^{{3 \over 2}}} + {3 \over 4}{x^{{4 \over 3}}} + C\)b) \(\eqalign{
& \int {{{x\sqrt x + \sqrt x } \over {{x^2}}}} dx = \int {{1 \over {\sqrt x }}} dx + \int {{{dx} \over {x\sqrt x }}} = \int {{x^{ - {1 \over 2}}}} dx + \int {{x^{ - {3 \over 2}}}} dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{{x^{{1 \over 2}}}} \over {{1 \over 2}}}\, + {{{x^{ - {1 \over 2}}}} \over {{-1 \over 2}}}\, + C = 2\sqrt x - {2 \over {\sqrt x }} + C \cr} \)c) \(\int {4{{\sin }^2}xdx = \int {2\left( {1 - \cos 2x} \right)dx} = 2\int {dx - 2\int {\cos 2xdx} } } = 2x - \sin 2x + C\)d) \(\int {{{1 + \cos 4x} \over 2}dx = {x \over 2}} + {1 \over 8}\sin 4x + C\)loigiaihay.com
