Question Detail

To evaluate ∫ x/(x-1) dx, we can use algebraic manipulation or a substitution.Let u = x - 1, then x = u + 1 and dx = du.The integral becomes ∫ (u + 1)/u du = ∫ (1 + 1/u) du.Integrating term by term: ∫ 1 du + ∫ 1/u du = u + ln|u| + C.Substitute back u = x - 1: (x - 1) + ln|x - 1| + C.Since C - 1 is still an arbitrary constant, we can write it as x + ln|x - 1| + C.Option B matches this form, assuming x-1 > 0 for the logarithm.