∫x/(x^4+2(x^2)+5)dx

来源：学生作业帮编辑：拍题作业网作业帮分类：数学作业时间：2024/05/29 20:02:41

∫x/(x^4+2(x^2)+5)dx

令a=x²
原式=1/2*∫dx²/(x^4+2x²+5)
=1/2*∫da/(a²+2a+1+4)
=1/2*∫da/[(a+1)²+4]
=1/8*∫da/[(1/4)(a+1)²+1]
=1/8*∫da/[(a/2+1/2)²+1]
=1/4*∫d(a/2+1/2)/[(a/2+1/2)²+1]
=1/4*arctan((a/2+1/2)+C
=1/4*arctan(x²/2+1/2)+C

∫2^x*3^x/(9^x-4^x) dx ∫x/（x^2+5）dx ∫（x^2+1/x^4)dx ∫dx/根号(4x-x^2) ∫1/(x^4-x^2)dx ∫(x^4-4x^2+5x-15)/(x^2+1)(x-2) dx=? ∫ [(x^3-2x^2+x+1)/(x^4+5x^2+4)]dx 求积分∫dx/(根号5-4x-x^2) ∫(3x-2)/(x^2-2x+5)dx ∫(x+1)/(x^2-2x+5)dx ∫[（x+1)/(x^(2)+2x+5]dx ∫（(x+2)/4x(x^2-1)）dx