∫1/x(x+1)∧3dx

来源：学生作业学帮网编辑：学帮网时间：2024/06/04 16:30:12

∫1/x(x+1)∧3dx

∫1/x(x+1)^3dx
=∫(x+1-x)/x(x+1)^3dx
=∫1/x(x+1)^2dx-∫1/(x+1)^3dx
=∫(x+1-x)/x(x+1)^2dx-∫1/(x+1)^3dx
=∫1/x(x+1)dx-∫1/(x+1)^2dx-∫1/(x+1)^3dx
=∫(x+1-x)/x(x+1)dx-∫1/(x+1)^2dx-∫1/(x+1)^3dx
=∫1/xdx-∫1/(x+1)dx-∫1/(x+1)^2dx-∫1/(x+1)^3dx
=∫1/xdx-∫1/(x+1)d(x+1)-∫1/(x+1)^2d(x+1)-∫1/(x+1)^3d(x+1)
到这里应该会做了吧.

∫e∧(-3x+1)dx ∫1/x(1+x^3)dx ∫x+1／(x-1)^3dx 求∫x/(1+x)^3 dx ∫ x/ (1+x)^3 dx ∫dx/x(x^3+1) ∫x^3/1+x^2 dx ∫(x-1)^2/x^3 dx ∫(X^3)/(1+X^2)dx ∫1/（x^3+x） dx ∫dx/x(1+x) x-9/[(根号)x]+3 dx ∫ x+1/[(根号)x] dx ∫ [（3-x^2）]^2 dx ∫1/x(x+1)∧3dx ∫0～∞x/(1+x)∧3dx= ∫（2-3x）³dx ∫1/（2x+5）∧11dx ∫x^3/(x^8-2) dx∫(x^3-1)/(x^2+1) dx ∫1/(x^100+x)dx ∫1/(e^x+e^3x)dx ∫(x^3-x^2+x+1)/(x^2+1) dx∫(x+4)/(x^2-x-2) dx