If \int {\log _e (x+\sqrt{1+x^2}) }/{\sqrt{1+x^2}} {\d} x= {\f}( {\g}(x))+ {\c} then

Correct Answer is : f(x)=x22,g(x)=loge(x+√1+x2)

{\f}(x)=\frac{x^2}{2}, {\g}(x)=\log _{ {\e}} (x+\sqrt{1+x^2})

{\f}(x)=\log _{ {\e}} (x+\sqrt{1+x^2}) , {\g}(x)=\frac{x^2}{2}

{\f}(x)=x^2, {\g}(x)=\log _{ {\e}} (x+\sqrt{1+x^2})

{\f}(x)=\log _{ {\e}} (x-\sqrt{1+x^2}) , {\g}(x)=x^2

WBJEE 2024 Math Paper

Question : 50 of 75

Marks: +1, -0

If ∫‌

loge(x+√1+x2)

√1+x2

dx=f(g(x))+c then

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