Question

二阶导数求解

二阶导数求解f"(x)+ax=0，求解

Answer 1

∫√（x^2-1）dx令x=sect 则 ∫√（x^2-1）dx=∫tantdsect=∫tan^2tsectdt=∫(sec^2t-1)sectdt=∫(sec^3t-sect)dt=tant*sect-∫sec^3tdt即∫(sec^3t-sect)dt=tant*sect-∫sec^3tdt2∫(sec^3t)dt=tant*sect+∫sectdt∫sec^3tdt=1/2tant*sect+1/2ln|sect+tant|+c所以 ∫√（x^2-1）dx=tant*sect-∫sec^3tdt=1/2tant*sect-1/2ln|sect+tant|+c=1/2x√（x^2-1）-1/2ln|x+√（x^2-1）|+c

Answer 2