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Evaluate int_(6)^(e^(3)+5)(3x-16)/(x-5)dx. Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

Evaluate ∫6e3+53x−16x−5dx \int_{6}^{e^{3}+5} \frac{3 x-16}{x-5} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).

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Evaluate definite integrals using the chain ruleright chevron icon

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Q. Evaluate ∫6e3+53x−16x−5dx \int_{6}^{e^{3}+5} \frac{3 x-16}{x-5} d x . Write your answer in simplest form with all logs condensed into a single logarithm (if necessary).
  1. Rewrite Integral: We have the integral: \newline∫6e3+53x−16x−5 dx\int_{6}^{e^{3}+5}\frac{3x-16}{x-5}\,dx\newlineTo solve this integral, we will use the method of partial fractions. However, we notice that the numerator is already a degree less than the denominator, so we can proceed directly to integration.
  2. Integrate Constant Term: We can rewrite the integral as:\newline∫6e3+5(3−3⋅5−16x−5)dx\int_{6}^{e^{3}+5}(3 - \frac{3\cdot 5-16}{x-5})dx\newlineThis simplifies to:\newline∫6e3+5(3−1x−5)dx\int_{6}^{e^{3}+5}(3 - \frac{1}{x-5})dx\newlineNow we can integrate term by term.
  3. Integrate Fraction Term: The integral of the constant 33 is:\newline∫6e3+53dx=3x∣6e3+5\int_{6}^{e^{3}+5}3dx = 3x \bigg|_{6}^{e^{3}+5}\newlineEvaluating this from 66 to e3+5e^{3}+5 gives:\newline3(e3+5)−3(6)=3e3+15−18=3e3−33(e^{3}+5) - 3(6) = 3e^{3} + 15 - 18 = 3e^{3} - 3
  4. Combine Results: The integral of −1/(x−5)-1/(x-5) is:\newline∫6e3+5(−1x−5)dx=−ln⁡∣x−5∣∣6e3+5\int_{6}^{e^{3}+5}\left(-\frac{1}{x-5}\right)dx = -\ln|x-5| \bigg|_{6}^{e^{3}+5}\newlineEvaluating this from 66 to e3+5e^{3}+5 gives:\newline−ln⁡∣e3+5−5∣+ln⁡∣6−5∣=−ln⁡∣e3∣+ln⁡∣1∣=−3ln⁡(e)+ln⁡(1)-\ln|e^{3}+5-5| + \ln|6-5| = -\ln|e^{3}| + \ln|1| = -3\ln(e) + \ln(1)\newlineSince ln⁡(e)=1\ln(e) = 1 and ln⁡(1)=0\ln(1) = 0, this simplifies to:\newline−3(1)+0=−3-3(1) + 0 = -3
  5. Combine Results: The integral of −1/(x−5)-1/(x-5) is:\newline∫6e3+5(−1x−5)dx=−ln⁡∣x−5∣∣6e3+5\int_{6}^{e^{3}+5} \left(-\frac{1}{x-5}\right)dx = -\ln|x-5| \bigg|_{6}^{e^{3}+5}\newlineEvaluating this from 66 to e3+5e^{3}+5 gives:\newline−ln⁡∣e3+5−5∣+ln⁡∣6−5∣=−ln⁡∣e3∣+ln⁡∣1∣=−3ln⁡(e)+ln⁡(1)-\ln|e^{3}+5-5| + \ln|6-5| = -\ln|e^{3}| + \ln|1| = -3\ln(e) + \ln(1)\newlineSince ln⁡(e)=1\ln(e) = 1 and ln⁡(1)=0\ln(1) = 0, this simplifies to:\newline−3(1)+0=−3-3(1) + 0 = -3Combining the results from the two integrals, we get:\newline(3e3−3)−3=3e3−3−3=3e3−6(3e^{3} - 3) - 3 = 3e^{3} - 3 - 3 = 3e^{3} - 6\newlineThis is the final answer in its simplest form.

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