ln∣x−2x+2∣+c\ln \left|\frac{x-2}{x+2}\right|+clnx+2x−2+c
1xln∣x−2x+2∣+c\frac{1}{x} \ln \left|\frac{x-2}{x+2}\right|+cx1lnx+2x−2+c
x−ln∣x−2x+2∣+cx-\ln \left|\frac{x-2}{x+2}\right|+cx−lnx+2x−2+c
x+ln∣x−2x+2∣+cx+\ln \left|\frac{x-2}{x+2}\right|+cx+lnx+2x−2+c