how do I slove this integral? its the integral of(1/3x-1 -1/3x+1)
Take the integral: integral 0 dx The integral of zero is an arbitrary constant of integration: Answer: | = 0+constant
This might be meant as: \(\int(\frac{1}{3x-1}-\frac{1}{3x+1})dx\)
If so then it results in: \(\frac{1}{3} \ln(3x-1)-\frac{1}{3} \ln(3x+1)\rightarrow \frac{1}{3} \ln (\frac{3x-1}{3x+1})\)
