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Q13E

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Essential Calculus: Early Transcendentals
Found in: Page 334
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Evaluate the integral: \(\int {\frac{{axdx}}{{{x^2} - bx}}} \)

Hence, the value of \(\int {\frac{{axdx}}{{{x^2} - b}}} \) is \(a\log \left| {x - b} \right| + c\)

See the step by step solution

Step by Step Solution

Step 1: Factorizing

\(\int {\frac{{axdx}}{{{x^2} - bx}}} = \int {\frac{{axdx}}{{x(x - b)}} = \int {\frac{{adx}}{{x - b}}} } \)

Step 2: Finding value

\(\int {\frac{{adx}}{{x - b}}} \)

Let \(x - b = v\)

\(\begin{array}{l}1 = \frac{{dv}}{{dx}}\\ \Rightarrow dx = dv\\ \Rightarrow a\int {\frac{{dv}}{v} = a\log \left| v \right|} + c\\ = a\log \left| {x - b} \right| + c\end{array}\)

Hence, the value of \(\int {\frac{{axdx}}{{{x^2} - b}}} \) is \(a\log \left| {x - b} \right| + c\)

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