The value of the integral ∫01/2 1+3(x+1)2(1−x)61/4dx is__.

The value of the integral 01/21+3(x+1)2(1x)61/4dx is__.

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    Solution:

    I=01/21+3(x+1)2(1x)61/4dx=01/2(1+3)dx(1+x)2(1x)6(1+x)61/4

    Put  1x1+x=t

     2dx(1+x)2=dt I=11/3(1+3)dt2t3/2=(1+3)2×2t11/3 =(1+3)(31)=2

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