在线等大神帮忙解答几个微积分题目，题目如附图所示，可以写好发照片给我。100分！及时且完善再追加100！ 求数学大神拯救一下妹纸的微积分~跪谢了TT-TT~戳我吧戳我...
\u6c42\u9ad8\u6570\u5927\u795e\u89e3\u7b54\u5fae\u79ef\u5206\u9898\u76ee\uff0c\u7b2c(3)\u9898\uff0c\u5728\u7ebf\u7b49\uff0c\u6025\uff0c\u53ef\u8ffd\u52a0\u60ac\u8d4f\uff01
\u4e0d\u61c2\u7684\u518d\u95ee\uff0c\u671b\u91c7\u7eb3
\u7b2c2\u9898\u544a\u8bc9\u51c6\u786e\u533a\u95f4\u7684\u8bdd\u53ef\u4ee5\u5e2e\u4f60\u89e3
∴∫(2x-1)/(x^2+3x+2)
=∫[-3/(x+1)+5/(x+2)]dx
=∫[-3/(x+1)]d(x+1)+∫[5/(x+2)]d(x+2)
=-3ln|x+1|+5ln|x+2|+C
2、利用变限积分求导公式：
d[∫<a,φ(x)>f(t)dt]/dx=f[φ(x)]*φ'(x)
d[∫<0,x^2>sintdt]/dx=sin(x^2)*2x=2xsin(x^2)
∴lim[(∫<0,x^2>sintdt)/x] x->0
=lim{d[∫<0,x^2>sintdt]/dx}/(dx/dx) 洛必达法则
=lim[2xsin(x^2)/1]
=0
3、∫<1,+∞>1/x^4*dx
=∫<1,+∞>x^(-4)dx
=<1,+∞>[-1/3*x^(-3)]
=-1/3-0
=-1/3
4、5两题用二元函数极限的洛必达法则
lim[f(x,y)/g(x,y)]=lim{[f'x(x,y)dx+f'y(x,y)dy]/[g'x(x,y)dx+g'y(x,y)dy]}
其中，极限点为 x->x0, y->y0，且dx=x-x0, dy=y-y0
第4题：lim{[√(xy+4)-2]/(xy)} x->1, y->0
f(x,y)=√(xy+4)-2, g(x,y)=xy; x0=1, y0=0; dx=x-1, dy=y
f'x(x,y)=1/2*y/√(xy+4), f'y(x,y)=1/2*x/√(xy+4)
g'x(x,y)=y, g'y(x,y)=x
∴lim{[√(xy+4)-2]/(xy)} x->1, y->0
=lim{[1/2*y/√(xy+4)*dx+1/2*x/√(xy+4)*dy]/[y*dx+x*dy]}
=lim[1/2*1/√(xy+4)]*{[y(x-1)+xy]/[y(x-1)+xy]}
=1/4*lim{[2xy-y]/[2xy-y]}
=1/4
第5题：lim[sin(xy)/y] x->1, y->0
f(x,y)=sin(xy), g(x,y)=y; x0=1, y0=0; dx=x-1, dy=y
f'x(x,y)=ycos(xy), f'y(x,y)=xcos(xy)
g'x(x,y)=0, g'y(x,y)=1
∴lim[sin(xy)/y] x->1, y->0
=lim{[ycos(xy)*dx+xcos(xy)*dy]/[0*dx+1*dy]}
=limcos(xy)*{[y(x-1)+xy]/[0+y]}
=lim1*{[2xy-y]/y}
=lim(2x-1)*lim(y/y)
=(2-1)*1
=1
吃饭上课。晚上做。
用图形来做，你懂的，自己动手印象深。
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