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Question

The value (s) of 10x4(1x)41+x2dx is (are)

A
227π
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B
2105
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C
0
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D
71153π2
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Solution

The correct option is A 227π
Let I=10x4(1x)41+x2dx
=10(x41)(1x)4+(1x)4(1+x2)dx
=10(x21)(1x)4dx+10(1+x22x)2(1+x2)dx
=10{(x21)(1x)4+(1+x2)4x+4x2(1+x2)}dx
=10((x21)(1x)4+(1+x2)4x+441x2)dx
=10(x64x5+5x44x2+441+x2)dx
=[x774x66+5x554x33+4x4tan1x]10
=1746+5543+44(π40)=227π

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