∫(0→2)(4-x^2)^1.5dx答案是3∏,怎么会有∏

∫（x^2+1/x^4)dx ∫dx/根号(4x-x^2) ∫1/(x^4-x^2)dx ∫x/(4+9x^2)dx 计算∫(0→2)|x^2-x|dx ∫(0→1) x[(2x-1)^8]dx 计算∫π/4~0 x/cos^2x dx ∫dx/根号x^2(4-X^2) 求导∫dx/ 根号x^2(4-X^2)DX ∫x-根号下x dx ∫lx-2l dx ∫1/根号下（4-x^2) dx ∫e^(-x) dx ∫2/根号下x dx ∫(1/x^2)sin(1/x) dx ∫(0→2)(4-x^2)^1.5dx答案是3∏,怎么会有∏ ∫ 4/(1-2x)^2 dx 求∫4/(x^2+4)dx ∫0~2π x|sinx|dx ∫(0 1)x(arctanx)^2dx ∫ 4/sin^2x dx ∫√(4-x^2)dx等于? 用换元法求 ∫dx/(4-9x^2) ∫√(4-x^2)dx