como:
∫7e2xdx=7/2e2x+C
∫4x2dx=4/3x3+C
∫2√xdx=∫2x1/2dx=4/3√x3
∫2/xdx=2lnx+C
entonces:
∫ [7e2x + 4x2 + 2√x + 2/x ]dx =∫7e2xdx + ∫4x2dx + ∫2√xdx + ∫2/xdx = 7/2e2x+ 4/3x3 + 4/3√x3+ 2lnx+C
por lo que:
∫12 [7e2x + 4x2 + 2√x + 2/x ]dx = [7/2e2x+ 4/3x3 + 4/3√x3+ 2lnx]12 = [7/2e2·2+ 4/3·23 + 4/3√23+ 2ln2]-[7/2e2·1+ 4/3·13 + 4/3√13+ 2ln1]=
=[7/2e4+ 32/3 + 8/3√2+ 2ln2]-[7/2e2+ 4/3 + 4/3] = 7/2e4- 7/2e2+ 8/3√2 + 2ln2 +8