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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 

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a
g(x)=tan−1⁡x,h(x)=log⁡|x|
b
g(x)=log⁡|x|,h(x)=tan−1⁡x
c
g(x)=h(x)=tan−1⁡x
d
g(x)=log⁡|x|,h(x)=log⁡|x|

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detailed solution

Correct option is D

Putting y2=f(x)=x+22x+3, we have x=3y2−21−2y2 and  dx=−2y1−2y22dy.So =−∫y⋅2y1−2y22⋅1−2y23y2−2dy=2∫y22y2−13y2−2dy=−2∫12y2−1−23y2−2dy=12log⁡1+2y1−2y−23log⁡3y+23y−2+CThus g(x)=log⁡|x| and h(x)=log⁡|x|.


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