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Question

If f(x)dx=Ψ(x), then x5f(x3)dx is equal to

A
13x3Ψ(x3)x2Ψ(x3)dx+C
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B
13[x3Ψ(x3)x2Ψ(x3)dx]+C
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C
13x3Ψ(x3)x3Ψ(x3)dx+C
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D
13x3Ψ(x3)3x3Ψ(x3)dx+C
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Solution

The correct option is A 13x3Ψ(x3)x2Ψ(x3)dx+C
x5f(x3)dx

x3=t3x2dx=dt

=tf(t)dt3=13[tf(t)dtΨ(t)dt]

=13[x3Ψ(x3)]Ψ(x3)x2dx+C

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