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Question

ex(x25x+8) dx =exf(x)+c then f(x)

A
x25x+12
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B
x2+7x+15
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C
x27x15
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D
x27x+15
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Solution

The correct option is C x27x+15
ex(x25x+8)dx
=exx2dx5ex.xdx+8ex dx
=x2ex=2xex dx5ex x+8exdx
x2ex7[xexex]+8ex
=x2ex7xex+15 e x
=en[x27x+15]+c
so, f(x) x27x+15

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