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Question

Let x>0 be a fixed real number. Then the integral 0et|xt|dt is equal to.

A
x+2ex1
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B
x2ex+1
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C
x+2ex+1
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D
x2ex+1
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Solution

The correct option is D x+2ex1
0er|(xt)|dt
=x0etII(xt)Idt+xetII(tx)Idt
{(xt)et(1)et}x0{(xt)et+et}x
=x+2ex1.

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