12. Techniques of Integration

Partial Fractions

12. Techniques of Integration

Partial Fractions

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Express the rational function as a sum or difference of two simpler fractions. Use a system of equations.
$\frac{1}{(2 x + 1) (2 x - 1)}$

$- \frac{1}{2 (2 x + 1)} + \frac{1}{2 (2 x - 1)}$

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

$\frac{1}{2 (2 x + 1)} + \frac{1}{2 (2 x - 1)}$

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

Verified step by step guidance

Step 1: Start by expressing the given rational function $ \frac{1}{(2x+1)(2x-1)} $ as a sum of two simpler fractions. Assume the form $ \frac{1}{(2x+1)(2x-1)} = \frac{A}{2x+1} + \frac{B}{2x-1} $, where $ A $ and $ B $ are constants to be determined.

Step 2: Multiply through by the common denominator $ (2x+1)(2x-1) $ to eliminate the denominators. This gives $ 1 = A(2x-1) + B(2x+1) $.

Step 3: Expand the terms on the right-hand side: $ 1 = A(2x) - A + B(2x) + B $, which simplifies to $ 1 = (2A + 2B)x + (-A + B) $.

Step 4: Equate the coefficients of like terms (coefficients of $ x $ and the constant term) on both sides of the equation. This gives the system of equations: $ 2A + 2B = 0 $ (coefficient of $ x $) and $ -A + B = 1 $ (constant term).

Step 5: Solve the system of equations to find $ A $ and $ B $. From $ 2A + 2B = 0 $, we get $ A + B = 0 $, so $ B = -A $. Substitute $ B = -A $ into $ -A + B = 1 $, which simplifies to $ -A - A = 1 $ or $ -2A = 1 $. Solve for $ A $ and $ B $, and substitute back into the partial fraction form.