12. Techniques of Integration

Partial Fractions

12. Techniques of Integration

Partial Fractions

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Multiple Choice

Express the rational function as a sum or difference or simpler rational expressions.
$\frac{13 x + 2}{2 x^{2} - x - 1}$

$\frac{13}{x - 1} + \frac{2}{2 - x}$

$\frac{13}{x + 1} - \frac{2}{2 - x}$

$\frac{5}{x - 1} + \frac{3}{2 x + 1}$

$\frac{5}{x + 1} + \frac{3}{2 x - 1}$

Verified step by step guidance

Factorize the denominator of the given rational function $ \frac{13x+2}{2x^2 - x - 1} $. Start by factoring $ 2x^2 - x - 1 $ into two binomials. Look for two numbers that multiply to $-2$ (the product of $2$ and $-1$) and add to $-1$ (the middle coefficient). This gives $ (2x + 1)(x - 1) $.

Rewrite the rational function as $ \frac{13x+2}{(2x+1)(x-1)} $. The goal is to express this as a sum of simpler fractions, $ \frac{A}{2x+1} + \frac{B}{x-1} $, where $A$ and $B$ are constants to be determined.

Set up the equation $ \frac{13x+2}{(2x+1)(x-1)} = \frac{A}{2x+1} + \frac{B}{x-1} $. Combine the right-hand side over a common denominator: $ \frac{A(x-1) + B(2x+1)}{(2x+1)(x-1)} $.

Equate the numerators: $ 13x + 2 = A(x-1) + B(2x+1) $. Expand the right-hand side: $ A(x-1) = Ax - A $ and $ B(2x+1) = 2Bx + B $. Combine terms: $ Ax - A + 2Bx + B = (A + 2B)x + (-A + B) $.

Match coefficients of $x$ and the constant terms from both sides of the equation. From $ 13x + 2 = (A + 2B)x + (-A + B) $, we get two equations: $ A + 2B = 13 $ (coefficient of $x$) and $ -A + B = 2 $ (constant term). Solve this system of equations to find $A$ and $B$. Substitute these values back into $ \frac{A}{2x+1} + \frac{B}{x-1} $ to get the final expression.