math

i) ∫[(e^[-x]) / (-3e^[-3x] - 3e^[-2x])] = -ln(3e^[-3x] + 3e^[-2x]) + c ii) ∫[(e^[-2x]) / (-3e^[-3x] - 3e^[-2x])] = -1/2 ln(3e^[-3x] + 3e^[-2x]) + c Explanation: These integrals can be evaluated by multiplying the numerator and denominator by the denominator's conjugate. This causes the denominator to become the difference of two squares, and the integral can be solved by taking the natural log of the denominator and adding a constant. Both answers still have the same denominator, however, the powers of e raised in the numerator of the second integral are one less than the first integral. This explains why the answer for the first integral is multiplied by -1, while the answer for the second integral is multiplied by -1/2.