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