The correct option is D g(x)=ln|x+√1+x2|
Let I=∫√x2+11−x2dx
⇒I=∫√x2+11−x2⋅√x2+1√x2+1dx=∫1+x2(1−x2)√1+x2dx=∫2−(1−x2)(1−x2)√1+x2dx=2∫1(1−x2)√1+x2dx−∫1√1+x2dx=2∫1(1−x2)√1+x2dx−ln|x+√1+x2|+C1
Now, let I1=2∫1(1−x2)√1+x2dx
Put x=tanθ⇒dx=(1+tan2θ)dθ
So, I1=2∫1+tan2θ(1−tan2θ)secθdθ
=2∫cosθcos2θdθ=2∫cosθ1−2sin2θdθ
Put sinθ=t⇒dt=cosθ dθ
So, I1=2∫11−2t2dt=1√2ln∣∣∣1+√2t1−√2t∣∣∣
Now, x=tanθ⇒sinθ=x√1+x2=t
So, I1=1√2ln∣∣∣√1+x2+√2x√1+x2−√2x∣∣∣
Hence, f(x)=ln∣∣∣√1+x2+√2x√1+x2−√2x∣∣∣ and g(x)=ln|x+√1+x2|