令x=2tanu则dx=2(secu)^2 du原式=∫1/[ (2tanu)^2* 2secu]* 2(secu)^2 du=1/2*∫secu/(tanu)^2 du=1/2*∫ cosu/sin^2 u du=1/2*∫ d(sinu)/(sinu)^2=-1/2* 1/sinu+C=-1/(2sinu)+C=-√(x^2+4)/x+C
求不定积分：∫ 1/[x^2*((x^2+4)^1/2)]
令x=2tanu则dx=2(secu)^2 du原式=∫1/[ (2tanu)^2* 2secu]* 2(secu)^2 du=1/2*∫secu/(tanu)^2 du=1/2*∫ cosu/sin^2 u du=1/2*∫ d(sinu)/(sinu)^2=-1/2* 1/sinu+C=-1/(2sinu)+C=-√(x^2+4)/x+C