If f(x)=x+22x+3. then ∫f(x)x21/2dxis equal to 12g1+2f(x)1−2f(x)−23h3f(x)+23f(x)−2 +C where 

If f(x)=x+22x+3. then f(x)x21/2dx

is equal to 12g1+2f(x)12f(x)23h3f(x)+23f(x)2 

+C where 

  1. A

    g(x)=tan1x,h(x)=log|x|

  2. B

    g(x)=log|x|,h(x)=tan1x

  3. C

    g(x)=h(x)=tan1x

  4. D

    g(x)=log|x|,h(x)=log|x|

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    Solution:

    Putting y2=f(x)=x+22x+3, we have 

    x=3y2212y2 and  dx=2y12y22dy.

    So =y2y12y2212y23y22dy=2y22y213y22dy=212y2123y22dy=12log1+2y12y23log3y+23y2+C

    Thus g(x)=log|x| and h(x)=log|x|.

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