The integral ∫2x−3×2+x+12 dx is equal to 

The integral 2x3x2+x+12 dx is equal to 

  1. A

    1x2+x+11633tan12x+13432x+1x2+x+1+C

  2. B

    1x2+x+11633tan12x-13+432x-1x2+x+1+C

  3. C

    12x2+x+1(2x+1)2x2+x+12+C

  4. D

    14x2+x+1+23tan1(2x+1)+C

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    Solution:

    We have, 

         I=2x3x2+x+12dx=(2x+1)4x2+x+12dxI=2x+1x2+x+12dx41(x+1/2)2+(3/2)22dx    I=1x2+x+14I1   where    I1=1[(x+1/2)2+(3/2)2]2dx 

    Putting x+1/2=(3/2) tanθ in I1We get

               I1=3/2sec2θdθ[(3/2)tanθ)]2+(3/2)22 I1=833cos2θdθ=8331+cos2θ2dθ I1=433{θ+12sin2θ}+C I1=433tan12x+13+122tanθ1+tan2θ+C I1=433tan12x+13+342x+1x2+x+1+C I=1x2+x+11633tan12x+13432x+1x2+x+1+C

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