Partial Fractions

Partial Fractions: Cracking Indefinite Integration for JEE Main

Hey there, future JEE toppers! Indefinite integration can seem like a maze, but trust me, it's a rewarding one. Partial fractions are your secret weapon for simplifying complex integrals, especially those involving rational functions. Get this topic right, and you'll see a definite boost in your JEE score!

What are Partial Fractions?

Imagine you have a complicated fraction, like $$\frac{5x + 2}{x^2 + 3x + 2}$$. Trying to integrate this directly? Ugh! Partial fraction decomposition allows us to break this down into simpler fractions that are much easier to integrate. Think of it like separating a mixed bag of candies into individual types to enjoy them better.

Conceptually, we're reversing the process of adding fractions with different denominators. We start with the result of that addition and work backward to find the original fractions.

1. Decomposing Rational Functions

A rational function is simply a fraction where both the numerator and denominator are polynomials. Partial fraction decomposition only works when the degree of the numerator is less than the degree of the denominator. If not, you'll need to perform polynomial long division first. This ensures we're dealing with a proper rational function.

2. Linear Factors

Let's start with the simplest case: when the denominator can be factored into distinct linear factors. A linear factor is something of the form $$(x - a)$$, where $$a$$ is a constant.

Here, $$P(x)$$ is the polynomial in the numerator, and $$(x-a)$$ and $$(x-b)$$ are the distinct linear factors in the denominator. Our goal is to find the constants $$A$$ and $$B$$.

How to find A and B?

1. Multiply both sides of the equation by the original denominator, $$(x-a)(x-b)$$. This will clear the fractions.
2. Substitute strategic values of $$x$$. Choose $$x = a$$ and $$x = b$$. This will make one of the terms zero, allowing you to solve for the other constant. For instance, if $$x = a$$, then the $$(x-a)$$ term becomes zero, and you can directly solve for $$B$$. Conversely, if $$x = b$$, then the $$(x-b)$$ term becomes zero, and you can directly solve for $$A$$.
3. Alternatively, you can expand both sides of the equation after clearing the denominators and equate coefficients of like powers of $$x$$. This will give you a system of linear equations that you can solve for $$A$$ and $$B$$.

Example: Let's decompose $$\frac{5x + 2}{(x+1)(x+2)}$$. We want to find $$A$$ and $$B$$ such that: $$\frac{5x + 2}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$. Multiply both sides by $$(x+1)(x+2)$$ to get $$5x + 2 = A(x+2) + B(x+1)$$. Let $$x = -1$$ to get $$-3 = A(1) + B(0)$$, so $$A = -3$$. Now let $$x = -2$$ to get $$-8 = A(0) + B(-1)$$, so $$B = 8$$. Thus, $$\frac{5x + 2}{(x+1)(x+2)} = \frac{-3}{x+1} + \frac{8}{x+2}$$. Now you can integrate each term easily!

3. Repeated Linear Factors

Now, let's deal with situations where a linear factor appears more than once in the denominator. For instance, $$(x-a)^2$$ or $$(x-a)^3$$.

Notice that we need a term for each power of the repeated factor, up to the highest power present in the denominator.

How to find A and B? The process is similar to the distinct linear factors case:

1. Multiply both sides by $$(x-a)^2$$ to clear the fraction.
2. Substitute $$x = a$$. This will directly give you the value of $$B$$.
3. To find $$A$$, you can either substitute another value of $$x$$ (other than $$a$$) or equate the coefficients of $$x$$ on both sides of the equation.

Example: Decompose $$\frac{x + 1}{(x-1)^2}$$. We want to find $$A$$ and $$B$$ such that $$\frac{x + 1}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$$. Multiply both sides by $$(x-1)^2$$ to get $$x + 1 = A(x-1) + B$$. Let $$x = 1$$ to get $$2 = A(0) + B$$, so $$B = 2$$. Now let's substitute $$x = 0$$: $$1 = A(-1) + 2$$, which gives us $$A = 1$$. Therefore, $$\frac{x + 1}{(x-1)^2} = \frac{1}{x-1} + \frac{2}{(x-1)^2}$$.

4. Quadratic Factors

Sometimes, the denominator will have a quadratic factor that cannot be factored further into real linear factors. This means the discriminant, $$b^2 - 4ac$$, is negative.

The key difference here is that the numerator above the irreducible quadratic factor is a linear expression, $$(Bx + C)$$, not just a constant.

How to find A, B, and C?

1. Multiply both sides by $$(x-a)(x^2+bx+c)$$ to clear the fractions.
2. Substitute $$x = a$$. This gives you the value of $$A$$.
3. Equate coefficients of $$x^2$$ and $$x$$ (or the constant term) on both sides. This will give you two equations with two unknowns (B and C), which you can solve simultaneously.

Example: Decompose $$\frac{x^2 + 1}{(x-1)(x^2 + x + 1)}$$. We want to find $$A$$, $$B$$, and $$C$$ such that $$\frac{x^2 + 1}{(x-1)(x^2 + x + 1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2 + x + 1}$$. Multiply by $$(x-1)(x^2 + x + 1)$$ to get $$x^2 + 1 = A(x^2 + x + 1) + (Bx+C)(x-1)$$. Let $$x = 1$$ to find $$2 = A(3)$$, thus $$A = \frac{2}{3}$$. Now, let's equate coefficients. For $$x^2$$: $$1 = A + B$$, so $$1 = \frac{2}{3} + B$$, which gives $$B = \frac{1}{3}$$. For the constant term: $$1 = A - C$$, so $$1 = \frac{2}{3} - C$$, which gives $$C = -\frac{1}{3}$$. Therefore, $$\frac{x^2 + 1}{(x-1)(x^2 + x + 1)} = \frac{2/3}{x-1} + \frac{(1/3)x - (1/3)}{x^2 + x + 1}$$.

Mastering partial fractions is crucial for tackling a wide range of integration problems in JEE Main. Keep practicing, and you'll be integrating like a pro in no time!