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Question

Evaluate 10x4(1x)41+x2dx

A
(π227)
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B
(227π)
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C
(722π)
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D
(227+π)
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Solution

The correct option is C (227π)
10x4(1x)41+x2dx=10x4[1+x22x]21+x2

==10x4(1+x2)dx410x5dx+410x61+x2dx

=[x25+x77]4[x66]10+410dx1+x2+401(x2+1)(x4+1x2)1+x2dx

=(15+17)4(16)4[tan1x]10+(15+113)

=227π

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