2022-04-05
综稿缺脊艺键渗节目
=√2/2+√2/2-1+√扮态2/2-1+√2/2
=(1/4)[ln|x-1|-ln√(x^2+1)-2arctanx-(x-1)/(x^2+1)]+C
=2(x+1)/x^2
=∫[1,x]lnu/(1+u^2)du=f(x)
=Lim[x×Sin[1/x^2],x-0]+Lim[((3-e^x)/(2+x))^cscx,x-0]
=lim(x)[tanx-sinx]/[xln(1+x)-x^2][√(1+tanx)+√(1+sinx)]
a+b=1,a-b=0,a=b=1/2,
2022-04-09
艺节告培
=lim(x)[sec^2x-cosx]/2[x/(1+x)+ln(1+x)-2x]
f(x) = 1/(1 + e^x),
cos√[x(1-cosx)] = 1 - (1/3)x^3 +o(x^3)
=lim2/3×3x/2x
y/(y+1)=dx/x,
=∫[1,1/x] -ln(1+u) / udu+ ∫[1,1/x] lnu / udu
=xtan(x/2)+2sin(x/2)cos(x/2)tan(x/2)-∫袜蔽唯tan(x/2)2cos^2(x/2)dx
=(1/√并纳2) ln|csc(x+π/4) - cot(x+π/4)| | (0->π/2)
=lim(x)[(1-cos^3(x))/cos^2(x)]/2[x/(1+x)+ln(1+x)-2x]