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Find the integral?

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2 Answers

∫ x(x2 + 2)2dx
 
Let u = x2 + 2
 
du = 2xdx
 
∫ x(x2 + 2)2dx = (1/2)∫u2du = (1/2)*(u3/3) = (1/6)*(x2 + 2)(evaluated between 0 and 1)
 
= (1/6)[(1)2 + 2]3 - (1/6)[(0)2 + 2]3 = (1/6)*(27 - 8) = (19/6)

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Alternatively or as a check:
f(x) = x(x^2+2)^2
f(x) = x(x^4+4x^2+4)
f(x) = x^5+4x^3+4x
One of f's antiderivatives is
F(x) = x^6/6+4x^4/4+4x^2/2
F(x) = x^6/6+x^4+2x^2
F(1)-F(0) = 1/6+1+2 = 19/6