y=xln(x+根号下x^2+1)(-无穷

令f(x)=y=xln[x+√(x²+1)]
那么f(-x)=-xln[-x+√(x²+1)]
=xln{1/[-x+√(x²+1)]
=xln{[x+√(x²+1)]/[x+√(x²+1)][-x+√(x²+1)]}
=xln{[x+√(x²+1)]/(-x²+x²+1)}
=xln[x+√(x²+1)]
=f(x)
所以是偶函数