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Question

a01a2+x2dx=

A
π/2
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B
π/3
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C
π/4
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D
π/4a
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Solution

The correct option is D π/4a

a01a2+x2dx

1a2+x2dx=1a2

d(xa)1+(xa)2=1ad(xa)1+(xa)2

=1atan1(xa)+c

a01a2+x2dx=[1atan1(xa)+c]a0

=[1atan1(aa)+c](0+c)
=1atan1(1)

=1aπ4

=π4a


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