∫x-1dx

∫dx/x(1+x)

∫[dx/(e^x(1+e^2x)]dx∫[1/(e^x(1+e^2x)]dx=-∫[1/((1+e^2x)]d(e^-x)=-arctan[e^(-x)]+C∫(0->2)(e^2x+1/x)dx=(1/2)e^2x+lnx：(0-

∫x(1+lnx)dx∫x(1+lnx)dx=∫(1+lnx)d(x²/2)=(1/2)x²(1+lnx)-(1/2)∫x²d(1+lnx)=x²/2+(1/2)x²lnx-(1/2)∫x&

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

∫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*(-1/u^2)du=∫sinudx^(-1)用分部积分法：∫sin1/xdx=∫sin

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

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

∫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=∫(-1,0)xe^(-x^2)dx+∫(0,1)xe^x^2dx=-1/2∫(-1,0)e^(-x^2)d(-x^2)+1/2∫(0,1)e^x^2dx^2=1/2

∫1/x(1+x^3)dx上下乘以X^2再积分

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

∫1/(x^2+x+1)dx∫1/(x²+x+1)dx=∫1/[(x+1/2)²+3/4]d(x+1/2)=(2/✔3)arctan[(2x+1)/✔3]+c公式∫1/(x²+a&#

∫1/[(√X)(1+X)]dx

∫x+1／(x-1)^3dxD

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

∫dx/x+√(1-x²)

∫(x+arctanx)/(1+x^2)dx∫(x+arctanx)/(1+x^2)dx令arctanx=u,则x=tanu,dx=sec²udu,代入原式得：∫(x+arctanx)/(1+x^2)dx=∫[(tanu+u)/(

∫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)dx=-1/x-0.5ln|1-x|+0.5ln|1+x|+C,C为常数

∫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)]letx^2=siny2xdx=cosydy∫dx/[x√(1-x^4)]=(1/2)∫(1/siny)dy=(1/2)ln|cscy-coty|+C=(1/2)ln|1/x^2-