∫1X³/1×dx

∫dx/x(1+x)∫dx/x(1+x)∫dx/x(1+x)

∫[dx/(e^x(1+e^2x)]dx∫[dx/(e^x(1+e^2x)]dx∫[dx/(e^x(1+e^2x)]dx∫[1/(e^x(1+e^2x)]dx=-∫[1/((1+e^2x)]d(e^-

∫x(1+lnx)dx∫x(1+lnx)dx∫x(1+lnx)dx∫x(1+lnx)dx=∫(1+lnx)d(x²/2)=(1/2)x²(1+lnx)-(1/2)∫x²d

∫dx/x(x2+1),∫dx/x(x2+1),∫dx/x(x2+1),令x=tant则dx=sec^2tdt于是∫dx/[x(x^2+1)]=∫sec^2t/[tantsec^2t]dt=∫dt/t

∫sin(1/x)dx∫sin(1/x)dx∫sin(1/x)dx令u=1/x,则du=-x^(-2)dx=-1/x^2dx,则dx=-x^2du=-1/u^2du∫sin1/xdx=∫sinu*(-

∫xln(1+x)dx∫xln(1+x)dx∫xln(1+x)dx原式=1/2∫ln(1+x)dx²=1/2x²ln(1+x)-1/2∫x²dln(1+x)=1/2x&s

∫(x-1)^2dx,∫(x-1)^2dx,∫(x-1)^2dx,∫（x-1）²dx=∫（x-1）²d（x-1）=1/3（x-1）³

∫x^1/2dx∫x^1/2dx∫x^1/2dx∫x^1/2dx=1/(1+1/2)x^(1+1/2)+c=2/3x^(3/2)+c

∫(1,-1)xe^(x|x|)dx∫(1,-1)xe^(x|x|)dx∫(1,-1)xe^(x|x|)dx∫(-1,1)xe^(x|x|)dx=∫(-1,0)xe^(-x^2)dx+∫(0,1)xe

∫1/x(1+x^3)dx∫1/x(1+x^3)dx∫1/x(1+x^3)dx上下乘以X^2再积分

∫1/x^2+x+1dx∫1/x^2+x+1dx∫1/x^2+x+1dx注：此题应该是“∫1/(x^2+x+1)dx”?若是,解法如下.原式=∫dx/[(x+1/2)²+(√3/2)&sup

∫1/(x^2+x+1)dx∫1/(x^2+x+1)dx∫1/(x^2+x+1)dx∫1/(x²+x+1)dx=∫1/[(x+1/2)²+3/4]d(x+1/2)=(2/d

∫1/[(√X)(1+X)]dx∫1/[(√X)(1+X)]dx∫1/[(√X)(1+X)]dx

∫x+1／(x-1)^3dx∫x+1／(x-1)^3dx∫x+1／(x-1)^3dxD

∫arcsin根号(x/1+x)dx∫arcsin根号(x/1+x)dx∫arcsin根号(x/1+x)dx分步积分得∫arcsin{[x/(1+x)]^(1/2)}dx=xarcsin{[x/(1+

∫dx/x+√(1-x²)∫dx/x+√(1-x²)∫dx/x+√(1-x²)

∫(x+arctanx)/(1+x^2)dx∫(x+arctanx)/(1+x^2)dx∫(x+arctanx)/(1+x^2)dx∫(x+arctanx)/(1+x^2)dx令arctanx=u,则

∫dx/x^2(1-x^2)∫dx/x^2(1-x^2)∫dx/x^2(1-x^2)∫dx/x^2(1-x^2)=∫1/x^2dx+∫1/(1-x^2)dx=-1/x+0.5*∫1/(1-x)+1/(

∫1/√x*(4-x)dx∫1/√x*(4-x)dx∫1/√x*(4-x)dxLog就是ln的意思.后面自己加一个常数C即可.

∫dx/[x√(1-x^4)]∫dx/[x√(1-x^4)]∫dx/[x√(1-x^4)]∫dx/[x√(1-x^4)]letx^2=siny2xdx=cosydy∫dx/[x√(1-x^4)]=(1