Calculus
posted by Anon .
Evaluate
integral of x times square root of (2x1)dx

x (2x1)^.5 dx
u = x
dv = (2x1)^.5 dx
du = dx
v = (1/3)(2x1)^1.5
so
(x/3)(2x1)^(1.5) (1/3) (2x1)^1.5 dx
(x/3)(2x1)^1.5(1/3)(1/5)(2x1)^2.5
(1/3)[ x(2x1)^1.5  (1/5)(2x1)^2.5 ]
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