Use integration by parts to evaluate the integral: [tex]$\int \ln (5 s-9) d s$[/tex]
Answer (2)
We use integration by parts to evaluate ∫ ln ( 5 s − 9 ) d s . By letting u = ln ( 5 s − 9 ) and d v = d s , we derive and simplify to obtain s ln ( 5 s − 9 ) − s − 5 9 ln ∣5 s − 9∣ + C as the final answer. The complete evaluation incorporates substitutions and the evaluation of simpler integrals.
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Apply integration by parts with u = ln ( 5 s − 9 ) and d v = d s .
Find d u = 5 s − 9 5 d s and v = s .
Use the integration by parts formula: ∫ u d v = uv − ∫ v d u to get s ln ( 5 s − 9 ) − 5 ∫ 5 s − 9 s d s .
Simplify and evaluate the remaining integral to obtain the final answer: s ln ( 5 s − 9 ) − s − 5 9 ln ∣5 s − 9∣ + C .
Explanation
Problem Analysis We are asked to evaluate the integral ∫ ln ( 5 s − 9 ) d s using integration by parts.
Choosing u and dv The integration by parts formula is ∫ u d v = uv − ∫ v d u . We choose u = ln ( 5 s − 9 ) and d v = d s .
Finding du and v Then we find d u and v . Differentiating u with respect to s , we get d u = 5 s − 9 5 d s . Integrating d v with respect to s , we get v = s .
Applying Integration by Parts Substituting into the integration by parts formula, we have: ∫ ln ( 5 s − 9 ) d s = s ln ( 5 s − 9 ) − ∫ s ⋅ 5 s − 9 5 d s = s ln ( 5 s − 9 ) − 5 ∫ 5 s − 9 s d s
Rewriting the Integrand Now we need to evaluate the integral ∫ 5 s − 9 s d s . We can rewrite the integrand as follows: 5 s − 9 s = 5 1 ⋅ 5 s − 9 5 s = 5 1 ⋅ 5 s − 9 5 s − 9 + 9 = 5 1 ⋅ ( 1 + 5 s − 9 9 )
Substituting Back into the Integral Substituting this back into our integral, we get: ∫ ln ( 5 s − 9 ) d s = s ln ( 5 s − 9 ) − 5 ∫ 5 1 ( 1 + 5 s − 9 9 ) d s = s ln ( 5 s − 9 ) − ∫ ( 1 + 5 s − 9 9 ) d s
Evaluating the Integral Now we evaluate the integral: ∫ ln ( 5 s − 9 ) d s = s ln ( 5 s − 9 ) − ∫ d s − 9 ∫ 5 s − 9 1 d s = s ln ( 5 s − 9 ) − s − 5 9 ln ∣5 s − 9∣ + C
Final Answer Therefore, the final result is: ∫ ln ( 5 s − 9 ) d s = s ln ( 5 s − 9 ) − s − 5 9 ln ∣5 s − 9∣ + C
Examples
Integration by parts is a powerful technique used in physics and engineering to solve problems involving products of functions. For example, when calculating the work done by a force that varies logarithmically with displacement, you might encounter an integral similar to the one we solved. Understanding how to apply integration by parts allows engineers to accurately determine energy expenditure in complex systems, ensuring efficient designs and reliable performance.
