# Calculus

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Evaluate
integral of x times square root of (2x-1)dx

• Calculus -

x (2x-1)^.5 dx

u = x
dv = (2x-1)^.5 dx

du = dx

v = (1/3)(2x-1)^1.5

so
(x/3)(2x-1)^(1.5) -(1/3) (2x-1)^1.5 dx

(x/3)(2x-1)^1.5-(1/3)(1/5)(2x-1)^2.5

(1/3)[ x(2x-1)^1.5 - (1/5)(2x-1)^2.5 ]

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