\int {\log (x+\sqrt{1+x^{2}})/\sqrt{1+x^{2}} d x is equal to

Correct Answer is : 12[log(x+√1+x2)]2+c

\frac{1}{2} \log (x+\sqrt{1+x^{2})+c

\frac{1}{2}[\log (x+\sqrt{1+x^{2})]^{2}+c