3.5 Q-4

Question Statement

Evaluate the integral:
$\int \frac{(a-b)x}{(x-a)(x-b)} , dx, \quad a > b$

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Background and Explanation

This is a rational function where the numerator and denominator are both linear functions of $x$. The integral requires partial fraction decomposition to split the fraction into simpler terms that can be integrated individually. A basic understanding of partial fractions and logarithmic integration is necessary.

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Solution

Step 1: Partial Fraction Decomposition

We start with: $\frac{(a-b)x}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$

Multiplying through by $(x-a)(x-b)$ eliminates the denominators: $(a-b)x = A(x-b) + B(x-a)$

Step 2: Solve for Coefficients

Expand and group terms: $(a-b)x = A(x) - Ab + B(x) - Ba$ $(a-b)x = (A + B)x - (Ab + Ba)$

Equating coefficients:

1. Coefficient of $x$: $A + B = a - b$
2. Constant term: $-Ab - Ba = 0$

Find $A$:

Substitute $x = b$ into the equation: $(a-b)b = B(b-a)$ $B = -b$

Find $B$:

Substitute $x = a$ into the equation: $(a-b)a = A(a-b)$ $A = a$

Step 3: Rewrite the Integral

Substituting $A = a$ and $B = -b$, the integral becomes: $\int \frac{(a-b)x}{(x-a)(x-b)} , dx = \int \frac{a}{x-a} , dx + \int \frac{-b}{x-b} , dx$

Step 4: Integrate

1. For the first term:
   $\int \frac{a}{x-a} , dx = a \ln|x-a|$
2. For the second term:
   $\int \frac{-b}{x-b} , dx = -b \ln|x-b|$

Combining the results: $\int \frac{(a-b)x}{(x-a)(x-b)} , dx = a \ln|x-a| - b \ln|x-b| + C$

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Key Formulas or Methods Used

1. Partial Fraction Decomposition:
   $\frac{P(x)}{Q(x)} = \frac{A}{\text{factor1}} + \frac{B}{\text{factor2}}$

2. Logarithmic Integration:
   $\int \frac{1}{x+c} , dx = \ln|x+c|$

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Summary of Steps

1. Set up partial fraction decomposition:
   $\frac{(a-b)x}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
2. Solve for $A$ and $B$:
   • $A = a$
   • $B = -b$
3. Rewrite the integral: $\int \frac{(a-b)x}{(x-a)(x-b)} , dx = \int \frac{a}{x-a} , dx + \int \frac{-b}{x-b} , dx$
4. Integrate: $a \ln|x-a| - b \ln|x-b| + C$
5. Final answer: $\boxed{a \ln|x-a| - b \ln|x-b| + C}$