1/(x+x^2)dx

来源：学生作业帮助网编辑：作业帮时间：2024/11/05 15:53:33

1/(x+x^2)dx
1/(x+x^2)dx

1/(x+x^2)dx
∫1/(x+x^2)dx
=∫1/xdx -∫1/(x+1)dx
=ln|x|-ln|x+1| +C

1/x+x^2=1/(x+1)x=1/x-1/1+x
所以原式等于lnx/1+x +C

∫1/(x+x^2)dx
=∫[1/x-1/(x+1)]dx
=∫1/xdx -∫1/(x+1)dx
=ln|x|-ln|x+1| +C

∫1/√(x^2-a^2) dx=|In(x+√(x^2-a^2))|+c
∫1/(x+x^2)dx
=∫1/[(x+1/2）^2-1/4)]d(x+1/2)
=∫1/√(x^2-a^2) dx=|In【x+1/2+√{(x+1/2)^2-1/4}】|+c

1/(x+x^2)dx ∫[dx/(e^x(1+e^2x)]dx ∫1+2x/x(1+x)*dx∫1+2x/x(1+x) * dx ∫(x-1)^2dx, ∫x^1/2dx x^2dx/(1-x^2)^2 微分方程.(1+x)^2/ (1-x)dx . ∫1/x^2+x+1dx ∫1/(x^2+x+1)dx ∫dx/x^2(1-x^2) ∫dx/x^2(1+x^2) ∫X^2/1-x^2 dx. dx/x^2(根号1+x^2) ∫X^2/1-x^2 dx. dx/(x+2)(2x+1) ∫ (x+arctanx)/(1+x^2) dx 不定积分dx/x(根号1-x^2) ∫（x^2+1/x^4)dx