回答如下:
∫1/(1-x^2)dx
=1/2∫[1/(1-x)+1/(1+x)]dx
=1/2[-ln(1-x)+ln(1+x)]+C
=1/2ln[(1+x)/(1-x)]+C
![](https://iknow-pic.cdn.bcebos.com/a8ec8a13632762d08fae332eb2ec08fa503dc68b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
不定积分的公式:
1、∫adx=ax+C,a和C都是常数
2、∫x^adx=[x^(a+1)]/(a+1)+C,其中a为常数且a≠-1
3、∫1/xdx=ln|x|+C
4、∫a^xdx=(1/lna)a^x+C,其中a>0且a≠1
5、∫e^xdx=e^x+C
6、∫cosxdx=sinx+C
7、∫sinxdx=-cosx+C
8、∫cotxdx=ln|sinx|+C=-ln|cscx|+C