Showing all your working, evaluate ∫ (21x^6 - e^2x- (1/x) +6)dx

When integrating a long chain of functions, we can integrate each term seperately and combine them. Let us now integrate:∫21x⁶dx = 21∫x⁶dx. Using the Power Rule [∫x^adx = (x^a+1/a+1)], we can say that 21∫x⁶dx = (21x⁷)/7 = 3x⁷. ∫e^2xdx. Now let u = 2x. du/dx = 2 so dx = du/2. Substitute both in to get:∫(e^u/2)du = 1/2∫e^udu. This is a common integral, which gives us 1/2 e^u = 1/2 e^2x. ∫(1/x)dx. This is a common integral which equals ln |x|∫6dx = 6∫dx = 6x (Integration of an integer).We then combine all the terms to give us 3x⁷ - e^2x/2 - ln |x| + 6x.When ever we integrate without limits, we have to add a constant c. This is unknown, unless addition information is given, so we call this C. Hence, the answer is:3x⁷ - e^2x/2 - ln |x| + 6x + C