∫(-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
設(shè)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挺簡(jiǎn)單的，統(tǒng)一積分區(qū)間，這樣可以合并兩個(gè)積分，往往可以使被積函數(shù)簡(jiǎn)化通過(guò)x=-y這個(gè)變換，把原積分區(qū)間(-2,0)變成(0,2)，兩個(gè)積分的積分區(qū)間就一樣了注意觀察對(duì)稱性，(-2,0)與(0,2)，這兩個(gè)區(qū)間關(guān)于原點(diǎn)對(duì)稱，因此考慮變換x=-y