smc

Factor out the $t^2$ inside the square root. $$\sqrt{t^2\cdot (t^2+1)}$$ $$\sqrt{t^2}\cdot \sqrt{t^2+1}$$ Remember that $\sqrt{t^2}=|t|$ because square rooting a nonnegative real number will always result in a nonnegative number. $$|t|\sqrt{t^2+1}$$ The answer is $\boxed{|t|\sqrt{1+t^2}}$.