解：f(x)=∫xln(1+x²)dx =(1/2)∫ln(1+x²)d(1+x²) =(1/2){(1+x²)ln(1+x²)-∫(1+x²)[(2x)/(1+x²)]dx} =(1/2)[(1+x²)ln(1+x²)-∫(2x)dx] =(1/2)[(1+x²)ln(1+x²)-x²]+C 代入点(0，-1/2)，C=-1/2. ∴f(x)=(1/2)(1+x²)[ln(1+x²)-1]
解：f(x)=∫xln(1+x²)dx =(1/2)∫ln(1+x²)d(1+x²) =(1/2){(1+x²)ln(1+x²)-∫(1+x²)[(2x)/(1+x²)]dx} =(1/2)[(1+x²)ln(1+x²)-∫(2x)dx] =(1/2)[(1+x²)ln(1+x²)-x²]+C 代入点(0，-1/2)，C=-1/2. ∴f(x)=(1/2)(1+x²)[ln(1+x²)-1]
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