1 / 3 [3Y - (10-7y / 2)] - 1 / 6 [2Y - (2Y + 2 / 3)] = Y / 2-1

1 / 3 [3Y - (10-7y / 2)] - 1 / 6 [2Y - (2Y + 2 / 3)] = Y / 2-1

Y-10/3+7/6Y-1/3Y+1/3Y+1/9=1/2Y-1
Y=10/9

How to calculate the indefinite integral ∫ x ^ 2 (2x ^ 3 + 1) DX

Note that the derivative of x ^ 3 is 3x ^ 2, so 1 / 3 D (x ^ 3) = x ^ 2 DX, so ∫ x ^ 2 (2x ^ 3 + 1) DX = 1 / 3 * ∫ (2x ^ 3 + 1) d (x ^ 3) and then add D (x ^ 3) to D (2x ^ 3 + 1), and multiply by 1 / 2 = 1 / 6 * ∫ (2x ^ 3 + 1) d (2x ^ 3 + 1) = 1 / 12 * (2x ^ 3 + 1) ^ 2 + C, C is a constant

Help indefinite integral calculation ∫ DX / (x ^ 2 √ (2X-4))

It should be like this: