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Question

If x2+11x2dx=12f(x)g(x)+C, where C is a constant of integration, then

A
f(x)=ln1+x2+2x1+x22x
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B
g(x)=ln|x1+x2|
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C
f(x)=ln1+x22x1+x2+2x
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D
g(x)=ln|x+1+x2|
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Solution

The correct option is D g(x)=ln|x+1+x2|
Let I=x2+11x2dx
I=x2+11x2x2+1x2+1dx=1+x2(1x2)1+x2dx=2(1x2)(1x2)1+x2dx=21(1x2)1+x2dx11+x2dx=21(1x2)1+x2dxln|x+1+x2|+C1

Now, let I1=21(1x2)1+x2dx
Put x=tanθdx=(1+tan2θ)dθ
So, I1=21+tan2θ(1tan2θ)secθdθ
=2cosθcos2θdθ=2cosθ12sin2θdθ

Put sinθ=tdt=cosθ dθ
So, I1=2112t2dt=12ln1+2t12t
Now, x=tanθsinθ=x1+x2=t
So, I1=12ln1+x2+2x1+x22x

Hence, f(x)=ln1+x2+2x1+x22x and g(x)=ln|x+1+x2|

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