已知∫xf(x)dx=e−x2+C\int_{x} f(x) dx = e^{-x^2}+C∫xf(x)dx=e−x2+C,则f(x)=f(x) =f(x)=( ).
(A)xe−x2xe^{-x^2}xe−x2
(B)−xe−x2-xe^{-x^2}−xe−x2
(C)2e−x22e^{-x^2}2e−x2
(D)−2xe−x2-2xe^{-x^2}−2xe−x2
∫1x+1dx=\int \dfrac{1}{\sqrt{x}+1} dx =∫x+11dx=( ).
已知limx→1f(x)=1\lim\limits_{x \to 1} f(x) = 1x→1limf(x)=1,limx→1g(x)=2\lim\limits_{x \to 1} g(x) = 2x→1limg(x)=2,则limx→1[f(x)+g(x)]=\lim\limits_{x \to 1} [f(x) + g(x)] =x→1lim[f(x)+g(x)]=( ).
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