If ∫ex2−x2(1−x)1−x2dx=μex1+x1−xλ+C then the value of  2(λ+μ) is 

If ex2x2(1x)1x2dx=μex1+x1xλ+C then the value of  2(λ+μ) is 

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    Solution:

    Let I=ex2x2(1x)1x2dx then

         I=ex1+1x2(1x)1x2dx    =ex1(1x)1x2+1+x1xdx     I=ex1+x1x+C     λ=12 and μ=1

    Hence,2(λ+μ)=3

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