3.5 Q-10

Question Statement

Evaluate the integral:
$\int \frac{2x-1}{x(x-1)(x-3)} dx$

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Background and Explanation

This problem requires knowledge of partial fraction decomposition. The fraction is broken into simpler terms that can be integrated individually. The concept of logarithmic integration is also used, as the resulting terms will involve $\ln$ functions.

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Solution

To simplify the given integral, we express the fraction as a sum of partial fractions:

$$\frac{2x-1}{x(x-1)(x-3)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-3}$$

Step 1: Multiply Through by the Denominator

Multiply both sides by $x(x-1)(x-3)$ to eliminate the denominators:

$$2x - 1 = A(x-1)(x-3) + Bx(x-3) + Cx(x-1)$$

Step 2: Solve for Coefficients ($A$, $B$, $C$)

Substitute convenient values for $x$ to isolate and solve for each coefficient.

When $x = 0$:

$$2(0) - 1 = A(0-1)(0-3)$$ $$-1 = A(-1)(-3) \quad \Rightarrow \quad A = \frac{-1}{3}$$

When $x = 1$:

$$2(1) - 1 = B(1)(1-3)$$ $$1 = B(-2) \quad \Rightarrow \quad B = \frac{-1}{2}$$

When $x = 3$:

$$2(3) - 1 = C(3)(3-1)$$ $$5 = 6C \quad \Rightarrow \quad C = \frac{5}{6}$$

Step 3: Rewrite the Integral

Substitute the values of $A$, $B$, and $C$ into the decomposition:

$$\frac{2x-1}{x(x-1)(x-3)} = \frac{-1}{3x} - \frac{1}{2(x-1)} + \frac{5}{6(x-3)}$$

Thus, the integral becomes:

$$\int \frac{2x-1}{x(x-1)(x-3)} dx = \int \left( \frac{-1}{3x} - \frac{1}{2(x-1)} + \frac{5}{6(x-3)} \right) dx$$

Step 4: Integrate Each Term

Using the formula $\int \frac{1}{u} du = \ln|u| + C$:

1. For $\frac{-1}{3x}$:
   $\int \frac{-1}{3x} dx = \frac{-1}{3} \ln|x|$
2. For $\frac{-1}{2(x-1)}$:
   $\int \frac{-1}{2(x-1)} dx = \frac{-1}{2} \ln|x-1|$
3. For $\frac{5}{6(x-3)}$:
   $\int \frac{5}{6(x-3)} dx = \frac{5}{6} \ln|x-3|$

Combine the results:

$$\int \frac{2x-1}{x(x-1)(x-3)} dx = \frac{-1}{3} \ln|x| - \frac{1}{2} \ln|x-1| + \frac{5}{6} \ln|x-3| + C$$

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Key Formulas or Methods Used

1. Partial Fraction Decomposition:
   $\frac{2x-1}{x(x-1)(x-3)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x-3}$
2. Logarithmic Integration:
   $\int \frac{1}{u} du = \ln|u| + C$

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Summary of Steps

1. Decompose the fraction into partial fractions:
   $\frac{2x-1}{x(x-1)(x-3)} = \frac{-1}{3x} - \frac{1}{2(x-1)} + \frac{5}{6(x-3)}$
2. Find coefficients $A$, $B$, and $C$ by substituting values of $x$.
   • $A = \frac{-1}{3}$, $B = \frac{-1}{2}$, $C = \frac{5}{6}$
3. Rewrite the integral and integrate each term individually.
4. Combine the results to get the final solution:
   $\int \frac{2x-1}{x(x-1)(x-3)} dx = \frac{-1}{3} \ln|x| - \frac{1}{2} \ln|x-1| + \frac{5}{6} \ln|x-3| + C$