Partial Fractions and Formulae -

What are Partial Fractions?

When integrating a rational function (a fraction where both numerator and denominator are polynomials), it can help to split it into simpler parts called partial fractions.
This is useful for solving integrals like $\int \frac{P(x)}{Q(x)} , dx$ , where $P(x)$ and $Q(x)$ are polynomials and the degree of $P(x)$ is less than the degree of $Q(x)$ .
The denominator $Q(x)$ should be factored into linear or quadratic terms (which cannot be factored further).

Steps for Using Partial Fractions

Factorize the denominator completely (into linear and/or irreducible quadratic factors).
Express the function as a sum of simpler rational functions (partial fractions) according to the types of factors present.
Find the unknown constants by equating and comparing coefficients or substituting suitable values of $x$ .
Integrate each simple fraction separately.

Types of Partial Fractions

Sr.	Form of Rational Function	Form of Partial Fraction
1	$\frac{px + q}{(x-a)(x-b)}$ $a \neq b$	$\frac{A}{x-a} + \frac{B}{x-b}$
2	$\frac{px + q}{(x-a)^2}$	$\frac{A}{x-a} + \frac{B}{(x-a)^2}$
3	$\frac{px^2 + qx + r}{(x-a)(x-b)(x-c)}$	$\frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}$
4	$\frac{px^2 + qx + r}{(x-a)^2(x-b)}$	$\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}$
5	$\frac{px^2 + qx + r}{(x-a)(x^2 + bx + c)}$ where $x^2 + bx + c$ cannot be factorised further	$\frac{A}{x-a} + \frac{Bx + C}{x^2 + bx + c}$

Notes

$A, B, C$ are real constants to be found by solving equations.
These forms cover most cases you will see in Class 11 and 12 for partial fractions.

Tip:
Whenever you see a rational function, first try to write it in one of the forms above. This will make integration much easier!

Example of Partial Fraction Decomposition

Express $\frac{2x + 3}{(x-1)(x+2)}$ as a sum of partial fractions.

Step 1: Identify the Form: The denominator is already factorized and matches the first row in the table:

$\frac{px + q}{(x-a)(x-b)}$

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

where $A$ and $B$ are constants to be found.

Step 2: Combine the Right Side: Combine the right side over a common denominator:

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

So,

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

Step 3: Equate the Numerators: Since denominators are equal, equate the numerators:

2x + 3 = A(x+2) + B(x-1)

Step 4: Expand and Group Like Terms: Expand the right side:

2x + 3 = Ax + 2A + Bx - B

Combine like terms:

2x + 3 = (A + B)x + (2A - B)

Step 5: Compare Coefficient: Match the coefficients of $x$ and the constant terms:

For $x: \quad 2 = A + B$
For constants: $3 = 2A - B$

Step 6: Solve for A and B

Now solve the two equations:

$A + B = 2$
$2A - B = 3$

Add the two equations:

(A + B) + (2A - B) = 2 + 3 \implies 3A = 5 \implies A = \frac{5}{3}

Substitute $A = \frac{5}{3}$ in the first equation:

\frac{5}{3} + B = 2 \implies B = 2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3}

Step 7: Write the Final Answer

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

Or, $\boxed{\frac{2x + 3}{(x-1)(x+2)} = \frac{5}{3(x-1)} + \frac{1}{3(x+2)}}$

Tip:
You can now integrate each term easily if required!

Example: Integration Using Partial Fractions

Evaluate: $\displaystyle \int \frac{2x + 3}{(x-1)(x+2)} , dx$

Step 1: Decompose into Partial Fractions

From our earlier solution,

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

Step 2: Write the Integral Using the Partial Fractions

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

Step 3: Take Constants Outside

= \frac{5}{3} \int \frac{1}{x-1} , dx + \frac{1}{3} \int \frac{1}{x+2} , dx

Step 4: Integrate Each Term

Recall, $\int \frac{1}{x-a} , dx = \ln|x-a| + C$

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

Step 5: Final Answer

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