text: calculus question? how would i solve this the integral of: the square root of (x) divided by the square root of [1+x(to the 3rd power)] dx put (1+x^3/2)=t 3/2x^1/2dx=dt x^1/2dx=(2/3)dt the integral=(2/3)dt/t =(2/3)log(1+x^3/2)+c