(根号下x的平方-1) 除以x的积分-搜答案-智慧作业帮

xdx/(1-x*x)^(1/2)=-1/2*d(1-x*x)/(1-x*x)^(1/2)再问：我也是这样算的最后是负一但答案是1

令x=3sect,则dx=secttantdt∫√(x^2-9)dx/x=∫tantsecttantdt/sect=∫(tant)^2dt=∫[(sect)^2-1]dt=tant-t+C=3/√(x

∫xdx/√(1-x²)=(1/2)∫2xdx/√(1-x²)=(1/2)∫dx²/√(1-x²)=-(1/2)∫d(-x²)/√(1-x²

x-1分之根号下x-1除以根号下x平方-x分之1=1/√（x-1)÷√（x³-1）/x=1/√（x²+x+1)/x=x/√（x²+x+1)∵x-1>0,∴x>1x=2原式

设x=sint,dx=costdt,(以下省略积分符号）原式=[(sint)^2/cost]costdt=(sint)^2dt=(1-cos2t)/2*dt=1/2[dt-cos2tdt)=1/2t-

-(根号2)/4利用分子有理化,分式上下同乘以(根号3-x加上根号1+x),得到2(1-x)/(x^2-1)(根号3-x加上根号1+x)=-2/(x+1)(根号3-x加上根号1+x)这时,可将x=1代

y=√(x^2+1)/(2x-1)y'=(1/2)*√(2x-1)/(x^2+1)*[(x^2+1)'(2x-1)-(x^2+1)(2x-1)']/(2x-1)^2=(1/2)*√(2x-1)/(x^

f(x)=(x+根号(1+x^2)/(1+x^2)=x/(1+x^2)+1/根号(1+x^2)f'(x)=[x'(1+x^2)-x*(1+x^2)']/(1+x^2)^2-1/2*(1+x^2)^(-

二分之根号2乘以arctan[(x-1)/根号(2x)]+四分之根号2乘以lnabs[(x+根号2x+1)/(x-2x+1)]+C

d[x^2/(x^2+5x)^(1/2)+x^3]={[2x(x^2+5x)^(1/2)-x^2(x^2+5x)^(-1/2)(2x+5)/2]/[x^2+5x]+3x^2}dx={[2(x^2+5x

1-√(1-4x²)=3x√(1-4x²)=1-3x平方1-4x²=1-6x+9x²13x²-6x=x(13x-6)=0x=0,x=6/13经检验,x

∫[dx(x^3)/√(1-x^2)]dx=-(1/3)(x^2+2)√(1-x^2)+C1分部积分,原式=∫arccosxd[-(1/3)(x^2+2)√(1-x^2)]=-(1/3)(x^2+2)

√（x-2)/（x-2)/√x/（x³-2x²）=√（x-2)²/（x-2)*（√（x²）/√x）=1*√x=√x

2x²+5>=0x-10解得x²>=-5/2x1所以x={x|x∈R,x≠1}再问：初二学生怎样写解题，x={x|x∈R,x≠1}这初中生不懂的再答：就是x≠1

2√(x²y)/3√(xy)=2√x√(xy)/3√(xy)=2(√x)/3

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

有理化.