Question

微积分初步题 求高手解答下 ，非常之感谢啊。。要给出基本步骤啊

Answer 1

1. lim [(x^2-2x-3)/(x^2-1)] = lim [(x+1)(x-3)/(x+1)(x-1)] = lim[x-3)/(x-1)] = lim[(-4)/(-2)] = 2

2. y = lnx + cos(e^x)
dy/dx = d(lnx)/dx + dcos(e^x)/dx let u = e^x, cos(e^x) = cos(u)
= 1/x + (dcosu/du)*(du/dx)
= 1/x + [-sin(u)*e^x]
= 1/x - (e^x)*sin(e^x)
dy = [1/x - (e^x)*sin(e^x)]dx
3. 积分符号打不出来, 用大写S 代替
Let u = e^(1/x), du/dx = -x(^-2)*e^(1/x), 所以 du = [-x(^-2)*e^(1/x)]dx
所以原式 = -Sdu = -S(1)du = -u + C = -e^(1/x) +C

4. Integral of lnx = xlnx - x
当 x = e, xlnx - x = e(lne) - e = 0,
当 x = 1, xlnx - x = (ln1) - 1 = -1,
所以原式 = 0 - (-1) = 1

Answer 2

1.分式上下同时除以x+1,再把x=-1代入就得到答案了。
2. dy=1/x-(e^x)*sin(e^x)
3.=-e^(1/x)+c
4=xlnx-x|=1

追问

能给点就算步骤吗，这样我没法交差啊，帮帮忙了，，，

追答

今天没带数据线，传不上图片。那些符号实在打不出来，抱歉了