求解高数题，请给出详细过程，谢谢
\u6c42\u89e3\u9ad8\u6570\u9898\uff0c\u8bf7\u7ed9\u51fa\u8be6\u7ec6\u8fc7\u7a0by''=2x(1-3x^2)/(1+x^2)^3
y''>0 => (-\u221e, -(\u6839\u53f73)/3),(0, (\u6839\u53f73)/3)\u4e0b\u51f8
y'' (-(\u6839\u53f73)/3, 0),((\u6839\u53f73)/3,+\u221e)\u4e0a\u51f8
y''=0, x=0, x=(\u6839\u53f73)/3, x=-(\u6839\u53f73)/3 \u9a8c\u8bc1\u540e\u5f97\u62d0\u70b93\u4e2a:(0, 0), ( -(\u6839\u53f73)/3, -(\u6839\u53f73)/4) ,( (\u6839\u53f73)/3, (\u6839\u53f73)/4)
==>(xy)'=e^x
==>xy=∫e^xdx=e^x+C (C是积分常数)
==>y=(e^x+C)/x
∴此方程的通解是y=(e^x+C)/x；
（2）原式=∫<0,1>dx∫<0,x^2>f(x,y)dy+∫<1,3>dx∫<0,(3-x)/2>f(x,y)dy。
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