Intergral

Brug substitution t = x³+2   dt = 3x²dx

   substitutionsmetoden

                                                u = x³+2

                                                du = 3x²dx

                                                x²dx = (1/3)du

                             ∫x² • (x³+2)⁵dx  =  ∫(x³+2)⁵·(x²dx)  = ∫u⁵·(1/3)du =  (1/3)·∫u⁵du = (1/3)·(1/6)u⁶ + k =

                                                             (1/18)·(x³+2)⁶ + k

Tusind tak!