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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
8105
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D
71153π2
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Solution

The correct option is C 227π

Expanding the numerator: x4(1x)4=x84x7+6x64x5+x4

Dividing this by 1+x2 by factorizing the terms and getting the reminder:

x4(1x)41+x2=x64x5+5x44x2+441+x2

I=10x4(1x)41+x2=10(x64x5+5x44x2+441+x2)

I=1/74/6+5/54/3+4/11041+x2

I=1/7+34tan11+4tan10

I=227π


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