∫1 1 根号1-x²

首先须满足2x-1>=0,即x>=1/2方程右边mx>=0,所以也有m>=0方程两边平方,因[x+√(2x-1)][x-√(2x-1)]=x^2-2x+1=(x-1)^2,得：2x+2|x-1|=m^

令√x=tx=t^2dx=2tdt原式=∫2tdt/(1+t)=2∫[1-1/(1+t)]dt=2t-2ln(1+t)+C

∫1/根号x*sec^2(1-根号x)dx=2∫sec^2(1-根号x)d(√x)=-2∫sec^2(1-根号x)d(1-√x)=-2tan(1-√x)+c

令t=√x∫1/(1+2√x)dx=∫1/(1+2t)dt^2=∫2t/(1+2t)dt=∫1-1/(1+2t)dt=∫dt-∫1/(1+2t)dt=t+1/2ln(1+2t)+C=√x+1/2ln(

y=根号（1-x^2）所以y^2=1-x^2即x^2+y^2=1又因为y=根号（1-x^2）大于等于0-1

解令√x=t则t²=x,dx=2tdt∴∫dx/(1+√x)=∫2tdt/(t+1)=2∫[(t+1)-1]/(t+1)dt=2∫1-1/(t+1)dt=2t-2ln|t+1|+C=2√x-

∫√[1+√x]/x^[3/4]dxLetu=x,dx=4udu=∫√[1+u]/u*[4u]du=4∫√[1+u]duLetu=tanz,du=seczdz=4∫√[1+tanz][seczdz]=

设√（5-4x）=yx=(5-y²)/4dx=-ydy/2则∫x/√(5-4x)dx=∫(5-y²)(-ydy/2)/4y=∫(y²-5)dy/8=y³/24-

设t=3次根号(x+1),x=t^3-1dx=3t^2dt原式=∫1/t*3t^2dt=∫3tdt=3/2t^2+C=3/2*3次根号(x+1)^2+C

∫dx/x根号(1+lnx)=∫1/根号(1+lnx)d(1+lnx)=2根号(1+lnx)＋c再问：=∫1/根号(1+lnx)d(1+lnx)为什么=2根号(1+lnx)＋c再答：∫dx/x根号(1

√x+1/√x=3,所以√x+1=3√x,1=2√x,所以√x=1/2.则√x-1/√x=(1/2-1)÷1/2=-1(负1)

∫1/[1+(√3x)]dx＝1/√3·∫1/[1+(√3x)]d(√3x)＝1/√3·∫1/[1+(√3x)]d(1+√3x)＝1/√3·ln|1+√3x|+C

∫x*√[(1-x)/(1+x)]dx=∫[x(1-x)/√(1-x^2)]dxletx=sinydy=cosydy∫[x(1-x)/√(1-x^2)]dx=∫siny(1-siny)dy=∫[sin

∫（（1+x-根号xcosx）根号x）dx=∫(√x+x√x-xcosx)dx=(2/3)x√x+(2/5)x²√x-∫xdsinx=2(x/3+x²/5)√x-xsinx+∫si

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

原方程根号内可配方：√[x+2-6√(x+2)+9]+√[x+2-10√(x+2)+25]=1√[√(x+2)-3]^2+√[√(x+2)-5]^2=1|√(x+2)-3|+|√(x+2)-5|=1记

[x+2√(x-1)]=[√(x-1)+1]^2[x-2√(x-1)]=[√(x-1)-1]^2x-1>=0x>=1y=√[x+2√(x-1)]+√[x-2√(x-1)]=√(x-1)+1+｜√(x-

∫(arctan√x)/[√x(1+x)]dx=∫(arctan√x)/(1+x)d(2√x)=2∫(arctan√x)/[1+(√x)²]d(√x)=2∫arctan√xd(arctan√

因为根号(-x^2)有意义,则x=0所以答案为1-4+0+2=-1