If ∫1×3+x4dx=Ax2+Bx+logx|x+1|+C, then 2A+2B=

If 1x3+x4dx=Ax2+Bx+logx|x+1|+C, then 2A+2B=

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    Solution:

    We have,

         I=1x3+x4dx=1x3(x+1)dx=1x2(x(x+1))dxI=1x21x1x+1dx=1x1x21x(x+1)dxI=1x1x21x1x+1dx=1x31x2+1x(x+1)dxI=1x31x2+1x1x+1dx=121x2+1x+logxx+1+C

    A=12 and B=12A+2B=1

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