∫1/(2+u^2) du= 1/√2 arctan u/√2?怎么来的
∫1/(2+u^2) du= 1/√2 arctan u/√2?怎么来的
∫du/(u^2-1)^(1/2)=ln[u+(u^2-1)^(1/2)]+C1
∫(u/(1+u-u^2-u^3)) du,求不定积分
∫(下限1上限1/x)[f(u)/u^2]du怎么求导
x=ln(u^2-1),dx={2u/(u^2-1)}du
请问∫sin(u/2)*sin(u/2)du或∫sin^2 (u/2)du怎么解啊?
原式=∫du/(1+u^2)(2u-1) =(-1/5)∫d(1+u^2)/(1+u^2)-(1/5)∫du(1+u^2
积分∫-4(u^2)/[(1-u^2)^2]du
求定积分∫(2-3)u^2/(u^2-1)du
求定积分∫(1,2) 2u/(1+u) du
求不定积分.∫【 u^(1/2)+1】(u-1) du:
du/(u^2-1)^(1/2)=dx/x 如何得到ln(u+(u^2-1))=lnx