Answer
$\ln x+\dfrac {\ln \left( 1+x^{2}\right) }{2}$
Work Step by Step
We use the rules of logarithms to obtain:
$\ln x\sqrt {1+x^{2}}=\ln x+\ln \sqrt {1+x^{2}}=\ln x+\ln \left( 1+x^{2}\right) ^{\frac {1}{2}}=\ln x+\dfrac {\ln \left( 1+x^{2}\right) }{2}$
