计算不定积分ln(1 根号下1 x x)dx-搜答案-智慧作业帮

∫ln(1-√x)dx=xln(1-√x)+(1/2)∫√x/(1-√x)dx=xln(1-√x)-(1/2)∫(1-√x-1)/(1-√x)dx=xln(1-√x)-(1/2)x+(1/2)∫1/(

表达式不够明确,可能被理解为两种情形：x^2(√x)/(1-x)或(x^2)*√[x/(1-x)]；如是第一种情形积分：设t=√x,则dx=2tdt；∫[x^2(√x)/(1-x)]dx=∫[2t^6

设f(x)=sin²x,g(x)=ln[x+√(1+x²)]f(x)明显是偶函数g(-x)=ln[-x+√(1+x²)]=ln{[√(1+x²)-x]̶

原式=∫1/√(1-ln²x)*1/xdx=∫1/√(1-ln²x)dlnx=arcsin(lnx)+C

∫dx／[x√(4－ln²x)]＝∫dlnx／√(4－ln²x)＝∫dt／√(4－t²)＝∫d(t／2)／√[1－(t／2)²]＝∫dm／√(1－m²

ln(x+1)的不定积分?

原式=∫ln(x+1)d(x+1)=(x+1)ln(x+1)-∫(x+1)dln(x+1)=(x+1)ln(x+1)-∫(x+1)*1/(x+1)d(x+1)=(x+1)ln(x+1)-∫dx=(x+

∫ln(x+√(1+x^2))dx=xln(x+√(1+x^2)-∫xd(ln(x+√(1+x^2))=xln(x+√(1+x^2)-∫xdx/√(1+x^2)=xln(x+√(1+x^2)-(1/2

用分步积分法∫ln(x+1)/√xdx=2∫ln(x+1)d√x=2ln(x+1)*√x-2∫√xdln(x+1)=2ln(x+1)*√x-2∫√x/(x+1)dx对于∫√x/(x+1)dx令√x=t

∫ln(x+√(1+x^2))dx=xln(x+√(1+x^2)-∫xd(ln(x+√(1+x^2))[ln(x+√1+x^2)]'=[1+x/√(1+x^2)]/(x+√(1+x^2))=1/√(1

(sqrt(x+1)*((4*x+1)*log(sqrt(x+1)+sqrt(x))-log(sqrt(x+1)-sqrt(x))-2*x*log(x))+sqrt(x)*((4*x+1)*log(s

原式=∫√[ln(x+√(1+x^2))+5]d(ln(x+√(1+x^2)+5)=1/2*[ln(x+√(1+x^2))+5]^(-1/2)+C

∫dx/x[根号1-(ln^2)x]=∫d(lnx)/[根号1-(ln^2)x]=∫dt/[根号1-t^2](设t=lnx)=arcsint+C=arcsin(lnx)+C

∫1/[x√(1-ln²x)]dx=∫1/√(1-ln²x)d(lnx)=arcsin(lnx)+C公式：∫dx/√(a²-x²)=arcsin(x/a)+C

可以用换元法解此题.令x=t^6则有原式=∫6t^5/(t^3+t^2)dt=∫6t^3/(t+1)dt然后将t^3分解为t+1的多项式,求出积分后将X=t^6代入则得结果

原式=1/2∫ln(x+1)dx²=1/2*x²ln(x+1)-1/2∫x²dln(x+1)=1/2*x²ln(x+1)-1/2∫x²/(x+1)dx

令x^(1/6)=u,则x=u^6,dx=6u^5du,√x=u³,x^(1/3)=u²∫1/[x^(1/2)-x^(1/3)]dx=∫6u^5/(u³-u²)

令x=cost,则dx=-sintdt∫√(1-x^2)/x^2dx=∫sint/(cost)^2·(-sint)dt=-∫(tant)^2dt=-∫[(sect)^2-1]dt=-∫(sect)^2