• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q14E

Expert-verified
Found in: Page 334

### Essential Calculus: Early Transcendentals

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

# 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-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 $$\frac{1}{{(a - b)}}\left( {\log \left( {\frac{{x + b}}{{x + a}}} \right)} \right) + c$$