∫(-2→2)x*ln(1+e^x)dx =∫(-2→0)x*ln(1+e^x)dx +∫(0→2)x*ln(1+e^x)dx ∫(-2→0)x*ln(1+e^x)dx 设y=-x,x=-y 原式=∫(2→0)(-y)*ln[1+e^(-y)]d(-y) =∫(2→0)y*ln[1+e^(-y)]dy =∫(2→0)y*ln[(e^y+1)/e^y]dy =∫(2→0)y*ln(e^y+1)dy -∫(2→0)y*ln(e^y)dy =-∫(0→2)y*ln(1+e^y)dy +∫(0→2)y^2dy 即∫(-2→0)x*ln(1+e^x)dx=-∫(0→2)x*ln(1+e^x)dx +∫(0→2)x^2dx 故∫(-2→2)x*ln(1+e^x)dx =∫(-2→0)x*ln(1+e^x)dx +∫(0→2)x*ln(1+e^x)dx =-∫(0→2)x*ln(1+e^x)dx +∫(0→2)x^2dx +∫(0→2)x*ln(1+e^x)dx =∫(0→2)x^2dx =[x^3/3]|(0→2) =2^3/3 =8/3
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