根号下x² 1 x²求不定积分-搜答案-神马作文网

答：∫dx/[1+√(1-x^2)]设x=sint,-π/2

∫arctan√xdx=xarctan√x-∫x*1/[1+(√x)^2]*1/2*1/√xdx=xarctan√x-1/2*∫√x/(1+x)*dx（令√x=t,则x=t^2,dx=2tdt）=xa

1、∫dx/(1-x)^2=∫d(x-1)/(x-1)^2=-1/(x-1)+C2、∫x√(1+2x^2)dx=1/4*∫√(1+2x^2)d(1+2x^2)=1/4*2/3*(1+2x^2)^(3/

Sx*根号下(1+x^2)dx=1/2*S(1+x^2)^(1/2)*d(1+x^2)=1/3*(1+x^2)^(3/2)+c

∫1/[x(√(1+lnx)]dx=2∫d√(1+lnx)=2√(1+lnx)+C

∫√[(1-x)/（1+x)]dx=∫(1-x)/√(1-x^2)dx=∫1/√(1-x^2)-∫x/√(1-x^2)dx=arcsinx+1/2∫(1-x^2)^(-1/2)d(1-x^2)=arc

答：∫x/√(1+x^2)dx=(1/2)∫[1/√(1+x^2)]d(x^2)=(1/2)∫(1+x^2)^(-1/2)d(x^2+1)=√(1+x^2)+C

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答：∫{1/[x√(1-x^2)]}dx设x=sint,-π/2再问：倒数第二步是怎么得出的？再答：常用积分表中的公式

设√x=t,x=t^2,dx=2tdt,原式=∫sint*2tdt=2∫t*sintdt=2∫td(-cost)=-2tcost+2∫costdt=-2tcost+2sint+C=-2√xcos√x+

∫1/(1-x)^2dx=1/(1-x)+c指数函数求积分∫x^a=[1/(a+1)]*x^(a+1)+c∫x*√(1+2x^2)dx=(1/4)*∫(1+2x^2)^(1/2)*d(1+2x^2)=

被积函数[x/(1-x³)]^(½)的定义域为x

换元,将x换为sinx或cosx

另根号（1+x²）=t原式=∫根号（t²-1）dt另根号（t²-1）=tanβt=secβ则原式=∫1/cos³βdβ=∫tanβsecβ+secβdβ=sec