3道简单的高等数学题目，完成再追加加分数。
\u5982\u4e0b\u56fe\u7247\u4e09\u9053\u9ad8\u6570\u9898\uff0c\u89e3\u7b54\u5b8c\u6709\u8ffd\u52a0\u5206\u6570\u30026.y'=2/3*(x+1)^(2/3-1)(x-5)^2+(x+1)^(2/3)*2(x-5)
=2/3*(x+1)^(-1/3)(x-5)^2+2(x+1)^(2/3)*(x-5)
=2/3*(x+1)^(-1/3)(x^-10x+25+3x^-12x-15)
=(8x^-44x+20)/[3(x+1)^(1/3)]
\u6781\u503c\u70b9\u5c31\u662f8x^-44x+20=0\u7684\u70b9
x=5,x=1/2 x5\u65f6\uff0c\u51fd\u6570\u9012\u589e,\u5176\u4f59\u9012\u51cf
x=5\u662f\u6781\u5c0f\u503c\u70b9f(5)=0\uff0cx=1/2\u662f\u6781\u5927\u503c\u70b9f(1/2)=(3/2)^(2/3)\u00d781/4
2.1 \u8bbef(x)=xln(x+\u221a(1+x^2))-\u221a(1+x^2)+1,\u5bf9\u5176\u6c42\u5bfc\uff0c
\u5373\u53ef\u5f97\u51faf'(x)=ln(x+\u221a(1+x^2))\uff0c\u82e5x>0,\u90a3\u4e48f'(x)>0\uff0c\u53e6\u5916\u53ef\u6c42\u51fa\uff0cf(0)=0\uff0c\u6240\u4ee5f(x)\u5f53x>0\u65f6\uff0cf(x)\u662f\u9012\u589e\u7684\uff0c
f(x)>f(0)=0,\u5373\u4e0d\u7b49\u5f0f\u6210\u7acb\u3002
4.f(x)=2+(x-4)/4-(x-4)^/64+o((x-4)^)
o((x-4)^)\u662f\u76ae\u4e9a\u8bfa\u4f59\u9879
\u6709\u7591\u95ee\u8bf7\u6307\u51fa
t + x dt/dx = t² - t
x dt/dx = t² - 2t
dt / (t² -2t) = dx /x 利用 1/ (t² -2t) = 1/2 [1 / (t -2) - 1 / t]
dt / [1 / (t -2) - 1 / t] = 2 dx /x
Ln((t-2) / t) = Lnx² + LnC
(t-2) / t = Cx ²
t = -2 / (Cx² -1)
即 y / x = -2 / (Cx² -1)
y = -2x / (Cx² -1)
代入条件当x = 1时， y(1) = 1，得到C = -1
特解为：y = 2x / (x² +1)
第三道：求二重积分∫∫√x^2+y^2 dxdy，其中积分区域D={(x,y)|x^2+y^2≤2x，0≤y≤x}。
参考答案：10/9√2
D={(x,y)|x²+y²≤2x，0≤y≤x}
===> D={(x,y)|(x-1)² + y²≤1，0≤y≤x}
===>方法1：先x后y：D={(x,y)|(x = y →1+√(1 - y²)，0≤y≤1}
方法2：极坐标：r² = 2rcosθ；即： θ = 0→π/4；r = 0→2cosθ
原积分 = ∫∫√x^2+y^2 dxdy
=∫{θ = 0→π/4}∫{r = 0→2cosθ} √r² r drdθ
=∫{θ = 0→π/4}∫{r = 0→2cosθ} r² drdθ
=∫{θ = 0→π/4} 8cos³θ /3 dθ
=8/3∫{θ = 0→π/4} 1 - sin²θ dsinθ
=8/3 [ sinθ - sin³θ/3 ] {sinθ = 0→1/√2}
=8/3 [1/√2 - 1/(6√2)]
=8/3 [5/(6√2)]
=10/(9√2)
1、采用分段积分，从1/e到1积分，结果为1；加上从1到无穷积分（结果为1），最后结果为2
2、假设y/x=p，则dy/dx=x*dp/dx+p
原方程化为
dp/(p^2-2p)=dx/x
-1/2*(1/p)--1/2*(1/(p-2))*dp=lnx+C
p/(p+2)=C/x^2
带入p和初值就可以得到
y=2x/(1+x^2)
3、采用换元法，x=1+cosa，y=sina
1
∫|lnx| dx/x^2=-∫|lnx|d(1/x)=-|lnx|/x+∫1/xd|lnx|
∫1/xd|lnx|=∫1/x^2dx=-1/x x→+∞ -|lnx|/x=0 1/x=0
x<e,∫1/xd|lnx| =-∫1/x^2dx=1/x+C
∫[1/e,+∞]|lnx|dx/x^2=1/(1/e)+1/(1/e)=2e
2
dy/dx = (y/x)^2 -y/x
y/x=p,y=xp
dy=pdx+xdp
dy/dx=p+xdp/dx
p+xdp/dx=p^2-p
dp/(p^2-2p)=dx/x
[(p-2)-p]dp/(p^2-2p)=-dlnx^2
dp/p-dp/(p-2)=-dlnx^2
dlnp-ln(p-2)=-dlnx^2
p/(p-2)=C/x^2
y/(y-2x)=C/x^2
y(1)=1
C=-1
(y-2x)/y=-x^2
1-2x/y=-x^2
1+x^2=2x/y
y=2x/(1+x^2)
kanbudonga
绛旓細绗簩閬擄細 浠 / x锛漷 ===> y=xt ===> dy/dx = t + x dt/dx t + x dt/dx = t² - t x dt/dx = t² - 2t dt / (t² -2t) = dx /x 鍒╃敤 1/ (t² -2t) = 1/2 [1 / (t -2) - 1 / t]dt / [1 / (t -2) - 1 / t...
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绛旓細鎴戠殑绛旀瑙佸浘鐗
