du/(u^2-1)=-tanh^-1(u)+c怎麼变的?

∫du/(u^2-1)^(1/2)=ln[u+(u^2-1)^(1/2)]+C1 x=ln(u^2-1),dx={2u/(u^2-1)}du ∫(u/(1+u-u^2-u^3)) du,求不定积分 ∫1/(2+u^2) du= 1/√2 arctan u/√2?怎么来的 du/(u^2-1)^(1/2)=dx/x 如何得到ln(u+(u^2-1))=lnx du=2/u^2+1 u(0)=5 自变量为t 用matlab 解次微分方程,且画出u随t的变化, matlab du/dt=d(du)/dx^2 x属于（0,1）,t属于(0,T]u(0,t)=u(1,t)=0u(x, 原式=∫du/(1+u^2)(2u-1) =(-1/5)∫d(1+u^2)/(1+u^2)-(1/5)∫du(1+u^2 (u/(1+u))du怎样积分成u-ln(u+1)? V=(v'+u)/{1+[(v*u)/(c^2)] } 求不定积分.∫【 u^（1/2）+1】（u-1) du: 求定积分∫(1,2) 2u/(1+u) du