5.2.3 Integrating with Partial Fractions

Introduction to Integration with Partial Fractions

Partial fractions allow us to decompose a complicated rational expression into simpler fractions that are easier to integrate. This method is particularly useful when the denominator can be factored into linear or irreducible quadratic terms. Once the fraction is decomposed, integration can proceed term by term.

Steps for Integration Using Partial Fractions

Decompose the fraction
Determine constants
Integrate each term

Decompose the fraction: Split the fraction into a sum of simpler fractions based on the factors in the denominator.

Determine constants: Solve for the unknown coefficients by equating numerators or substituting values.

Integrate each term: Use standard integration techniques for linear and quadratic terms.

Types of Partial Fraction Decomposition

Linear Factors Only:

If the denominator factors into distinct linear terms:

\frac{P(x)}{(x-a)(x-b)} \to \frac{A}{x-a} + \frac{B}{x-b}

Repeated Linear Factors:

For repeated linear factors, include terms for each power:

\frac{P(x)}{(x-a)^2} \to \frac{A}{x-a} + \frac{B}{(x-a)^2}

Irreducible Quadratic Factors:

For quadratic terms that cannot be factored further:

\frac{P(x)}{(x^2+1)} \to \frac{Ax+B}{x^2+1}

Worked Examples

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

Step 1: Decompose the fraction:

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

Step 2: Find $A$ and $B$ :

Multiply through by $(x-1)(x+2)$ and equate numerators:

1 = A(x+2) + B(x-1)

Expand:

1 = Ax + 2A + Bx - B

Group terms:

1 = (A+B)x + (2A-B)

Compare coefficients:

Coefficient of $x$ : $A + B = 0$

Constant term: $2A - B = 1$

Solve:

From $A+B=, B = -A$

Substitute into : $2A−B=1$

2A - (-A) = 1 \implies 3A = 1 \implies A = \frac{1}{3}

B = -A = -\frac{1}{3}

Step 3: Rewrite the fraction:

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

Step 4: Integrate term by term:

\int \frac{1}{(x-1)(x+2)} \, dx = \frac{1}{3} \int \frac{1}{x-1} \, dx - \frac{1}{3} \int \frac{1}{x+2} \, dx

= \frac{1}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C

Step 5: Simplify:

= \frac{1}{3} \ln\left(\frac{|x-1|}{|x+2|}\right) + C

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Example 2: Integrate $\frac{2x}{x^2+4}$

Step 1: Decompose the fraction:

The denominator $x^2+4$ is irreducible.

The partial fraction form is:

\frac{2x}{x^2+4} = \frac{A}{x^2+4}

Step 2: Simplify directly:

Observe that:

\int \frac{2x}{x^2+4} \, dx = \int \frac{d(x^2+4)}{x^2+4}

Step 3: Integrate:

Using the substitution $u = x^2+4, du = 2x \, dx$ :

\int \frac{2x}{x^2+4} \, dx = \int \frac{1}{u} \, du = \ln|u| + C = \ln(x^2+4) + C

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Example 3: Integrate $\frac{1}{x^3+2x}$

Step 1: Factorise the denominator:

x^3+2x = x(x^2+2)

Step 2: Decompose the fraction:

\frac{1}{x(x^2+2)} = \frac{A}{x} + \frac{Bx+C}{x^2+2}

Step 3: Find $A$ , $B$ , and $C$ :

Multiply through by $x(x^2+2)$ :

1 = A(x^2+2) + (Bx+C)x

Expand:

1 = Ax^2 + 2A + Bx^2 + Cx

Group terms:

1 = (A+B)x^2 + Cx + 2A

Compare coefficients:

Coefficient of $x^2$ : $A + B = 0$

Coefficient of $x$ : $C = 0$

Constant term: $2A = 1$

Solve:

From $2A = 1, A = \frac{1}{2}$

From $A + B = 0, B = -\frac{1}{2}$

$C = 0$

Step 4: Rewrite the fraction:

\frac{1}{x(x^2+2)} = \frac{\frac{1}{2}}{x} - \frac{\frac{1}{2}x}{x^2+2}

Step 5: Integrate term by term:

\int \frac{1}{x(x^2+2)} \, dx = \frac{1}{2} \int \frac{1}{x} \, dx - \frac{1}{2} \int \frac{x}{x^2+2} \, dx

First term:

\frac{1}{2} \int \frac{1}{x} \, dx = \frac{1}{2} \ln|x|

Second term:

Use substitution $u = x^2+2, u = 2x \, dx$

-\frac{1}{2} \int \frac{x}{x^2+2} \, dx = -\frac{1}{4} \ln(x^2+2)

Step 6: Combine results:

\int \frac{1}{x(x^2+2)} \, dx = \frac{1}{2} \ln|x| - \frac{1}{4} \ln(x^2+2) + C

Note Summary

infoNote

Common Mistakes:

Incorrect decomposition: Ensure the denominator is properly factored before setting up partial fractions.
Forgetting to solve for constants: Always solve for $A$ , $B$ , and $C$ accurately by equating numerators or substituting values.
Ignoring irreducible quadratic terms: For $ax^2+bx+c$ , include terms $\frac{Ax+B}{ax^2+bx+c}$
Not simplifying integrals: Some terms may directly simplify or use standard substitutions.
Missing the $+C$ : Always include the constant of integration in indefinite integrals.

infoNote

Key Formulas:

Linear factors:

\frac{P(x)}{(x-a)(x-b)} \to \frac{A}{x-a} + \frac{B}{x-b}

Repeated linear factors:

\frac{P(x)}{(x-a)^n} \to \frac{A}{x-a} + \frac{B}{(x-a)^2} + \dots + \frac{Z}{(x-a)^n}

Quadratic factors:

\frac{P(x)}{(ax^2+bx+c)} \to \frac{Ax+B}{ax^2+bx+c}

Standard integral results:

$\int \frac{1}{x} \, dx = \ln|x| + C$
$\int \frac{1}{x^2+a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$
$\int \frac{x}{x^2+a^2} \, dx = \frac{1}{2} \ln(x^2+a^2) + C$

Integrating with Partial Fractions Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.2.3 Integrating with Partial Fractions

Introduction to Integration with Partial Fractions

Steps for Integration Using Partial Fractions

Types of Partial Fraction Decomposition

Linear Factors Only:

Repeated Linear Factors:

Irreducible Quadratic Factors:

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

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