Calculus integration square root help?

Question

integrate the expression
____
V8X +1 dx (square root of 8X + 1 dx)

Answer 1

Let u = 8x+1 and du = 8 dx.
Thus (1/8)int(8sqrt(8x+1)dx) = (1/8)int(sqrt(u)du)
The integral of sqrt(u) is simply equal to (2/3)u^(3/2)
Thus you result with (1/12)u^(3/2)+C. You can substitue 8x+1 for u and get (1/12)*(8x+1)^(3/2)+c

Answer 2

Let u = 8x + 1, so du/8 = dx. The integrand becomes
(u^(1/2))du/8. Integrate by the power rule: the antiderivative is (1/8)*(u^(3/2))*(2/3) + C. The reverse substitution gives you
(1/12)*[(8x+1)^(3/2)] + C.