całka z egzaminu na szóstkę politechnika wrocławska Piotrek: ∫ (x⁶ + x³)³√x³ + 2dx

coped-zadania: integral ((x⁶+x³) 3 sqrt(x 3)+2) dx = 2/5 sqrt(3) x⁽15/2)+(2 x⁽9/2))/sqrt(3)+2 x+constant Possible intermediate steps: integral (2+3 sqrt(3) sqrt(x) (x³+x⁶)) dx Integrate the sum term by term and factor out constants: = 3 sqrt(3) integral sqrt(x) (x⁶+x³) dx+ integral 2 dx For the integrand sqrt(x) (x⁶+x³), substitute u = sqrt(x) and du = 1/(2 sqrt(x)) dx: = 6 sqrt(3) integral u² (u¹2+u⁶) du+ integral 2 dx Expanding the integrand u² (u¹2+u⁶) gives u¹4+u⁸: = 6 sqrt(3) integral (u¹4+u⁸) du+ integral 2 dx Integrate the sum term by term: = 6 sqrt(3) integral u¹4 du+6 sqrt(3) integral u⁸ du+ integral 2 dx The integral of u⁸ is u⁹/9: = 6 sqrt(3) integral u¹4 du+(2 u⁹)/sqrt(3)+ integral 2 dx The integral of u¹4 is u¹5/15: = (2 sqrt(3) u¹5)/5+(2 u⁹)/sqrt(3)+ integral 2 dx The integral of 2 is 2 x: = (2 sqrt(3) u¹5)/5+(2 u⁹)/sqrt(3)+2 x+constant Substitute back for u = sqrt(x): = 2/5 sqrt(3) x⁽15/2)+(2 x⁽9/2))/sqrt(3)+2 x+constant Jeśli nie rozumiesz − przetłumacz.