Question

二重积分求解

∫∫[x/(y+1)]dσ 其中D：y=x^2+1 y=2x x=0所围成的区域

Answer 1

求y=x^2+1与y=2x的交点：(1,2)先积y，y：2x--->x²+1，x：0--->1∫∫[x/(y+1)]dσ=∫[0--->1]∫[2x--->x²+1] [x/(y+1)]dydx=∫[0--->1] xln(y+1)| [2x--->x²+1] dx=∫[0--->1] [xln(x²+2)-xln(2x+1) ]dx以下略去积分限=∫xln(x²+2)dx-∫xln(2x+1)dx=1/2∫ln(x²+2)d(x²)-1/2∫ln(2x+1)d(x²)=1/2x²ln(x²+2)-1/2∫ 2x³/(x²+2)dx-1/2x²ln(2x+1)+∫x²/(2x+1)dx [0--->1]=(ln3)/2-∫ (x³+2x-2x)/(x²+2)dx-(ln3)/2+1/4∫(4x²-1+1)/(2x+1)dx=-∫xdx+∫ 2x/(x²+2)dx+1/4∫(2x-1)dx+1/4∫1/(2x+1)dx=-1/2x²+∫ 1/(x²+2)d(x²)+1/4x²-1/4x+1/8ln(2x+1) =-1/2x²+ln(x²+2)+1/4x²-1/4x+1/8ln(2x+1) [0--->1]=-1/2+ln3+1/4-1/4+(ln3)/8-ln2=(9ln3)/8-1/2-ln2数学软件验算：结果正确

Answer 2

二重积分的计算方法

Answer 3