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Evaluate int_(2)^(e+1)(4x-3)/(x-1)dx. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

Evaluate ∫2e+14x−3x−1dx \int_{2}^{e+1} \frac{4 x-3}{x-1} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

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Q. Evaluate ∫2e+14x−3x−1dx \int_{2}^{e+1} \frac{4 x-3}{x-1} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
  1. Antiderivative Calculation: The integral of a constant 44 with respect to xx is simply 4x4x. The integral of −1/(x−1)-1/(x-1) with respect to xx is −ln⁡∣x−1∣-\ln|x-1|. So we have:\newline∫(4x−3)/(x−1)dx=∫4dx−∫1/(x−1)dx=4x−ln⁡∣x−1∣+C\int(4x-3)/(x-1)dx = \int 4dx - \int 1/(x-1)dx = 4x - \ln|x-1| + C\newlinewhere CC is the constant of integration.
  2. Definite Integral Evaluation: Now we need to evaluate the definite integral from 22 to e+1e+1. We substitute the upper and lower limits into the antiderivative:\newline∫2e+14x−3x−1dx=[4x−ln⁡∣x−1∣]∣2e+1\int_{2}^{e+1}\frac{4x-3}{x-1}dx = [4x - \ln|x-1|] \Big|_{2}^{e+1}\newline= [4(e+1)−ln⁡∣e+1−1∣]−[4(2)−ln⁡∣2−1∣][4(e+1) - \ln|e+1-1|] - [4(2) - \ln|2-1|]\newline= [4e+4−ln⁡∣e∣]−[8−ln⁡∣1∣][4e+4 - \ln|e|] - [8 - \ln|1|]
  3. Final Simplification: We simplify the expression by substituting the natural logarithm of ee, which is 11, and the natural logarithm of 11, which is 00:[4e+4−ln⁡∣e∣]−[8−ln⁡∣1∣]=[4e+4−1]−[8−0]\left[4e+4 - \ln|e|\right] - \left[8 - \ln|1|\right] = \left[4e+4 - 1\right] - \left[8 - 0\right]=4e+4−1−8= 4e + 4 - 1 - 8=4e−5= 4e - 5

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