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Question

The value of the integral 10xdx(1+x)(1+3x)(3+x) is

A
π4(136)
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B
π8(132)
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C
π4(132)
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D
π8(136)
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Solution

The correct option is B π8(132)
Put x=t2
102t2(1+t2)(1+3t2)(3+t2)dt
2t2(1+t2)(1+3t2)(3+t2)=A1+t2+B1+3t2+C3+t2
A=12,B=38,C=38
1210dt1+t23810dt1+3t23810dt3+t2
=π838(13π3)38(13π6)
=π8316π=π8(132)

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