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Q14E

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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 \(\int {\frac{1}{{(x + a)(x + b)}}} dx\)

Hence, the value of integral \(\int {\frac{1}{{(x + a)(x + b)}}} dx\) is \(\frac{1}{{(a - b)}}\left( {\log \left( {\frac{{x + b}}{{x + a}}} \right)} \right) + c\)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step-1:  Using partial fraction,

\(\frac{1}{{(x + a)(x + b)}} = \frac{A}{{x + a}} + \frac{B}{{x + b}}\)

Step-2:  Finding value

\(\begin{array}{l}\frac{1}{{(x + a)(x + b)}} = \frac{{A(x + b) + B(x + a)}}{{(x + a)(x + b)}}\\1 = A(x + b) + B(x + a)\\1 = Ax + Ab + Bx + Ba\\1 = (A + B)x + Ab + Ba\end{array}\)

Step-3:  Comparing both the sides

\(\begin{array}{l}A + B = 0\\Ab + Ba = 1 \to (1)\\A + B = 0 \Rightarrow B = - A\end{array}\)

Putting in (1)

\(\begin{array}{l}Ab + Ba = 1\\Ab + ( - A)a = 1\\Ab - Aa = 1\\A(b - a) = 1 \Rightarrow A = \frac{1}{{b - a}}\end{array}\)

And \(B = - A = \frac{1}{{a - b}}\)

Step-4:  Finalizing the problem

\(\begin{array}{l}\frac{A}{{x + a}} + \frac{B}{{x + b}} = \frac{1}{{(b - a)}}\frac{1}{{(x + a)}} + \frac{1}{{(a - b)}}\frac{1}{{(x - b)}}\\\frac{1}{{b - a}}\int {\frac{1}{{x + a}}dx + \frac{1}{{a - b}}\int {\frac{{dx}}{{x + b}}} } \end{array}\)

Solving them

Let \(x + a = y\) and \(x + b = u\)

\(1 = \frac{{dy}}{{dx}}\) and \(1 = \frac{{du}}{{dx}}\)

\(dx = dy\) \(dx = du\)

\(\begin{array}{l}\frac{1}{{b - a}}\int {\frac{{dy}}{y} + \frac{1}{{a - b}}\int {\frac{{du}}{u}} } \\ = \frac{1}{{b - a}}\log \left| y \right| + \frac{1}{{(a - b)}}\log \left| u \right| + c\\ = \frac{1}{{b - a}}\log \left| {x + a} \right| + \frac{1}{{a - b}}\log \left| {x + b} \right| + c\\ = \frac{1}{{(a - b)}}\left( { - \log \left| {x + a} \right| + \log \left| {x + b} \right|} \right) + c\\ = \frac{1}{{(a - b)}}\left( {\log \left( {\frac{{x + b}}{{x + a}}} \right)} \right) + c\end{array}\)

Hence, the value of the integral \(\int {\frac{1}{{(x + a)(x + b)}}} dx\) is \(\frac{1}{{(a - b)}}\left( {\log \left( {\frac{{x + b}}{{x + a}}} \right)} \right) + c\)

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