Integrals Ex 7.5 Class 12 Maths NCERT: Partial Fractions

Welcome, Class 12 students! In our journey through integral calculus, you've learned various methods like substitution and integration by parts. Now, we're going to tackle a special type of integrand: rational functions. When you encounter integrals of the form P(x)/Q(x), where P(x) and Q(x) are polynomials, direct integration can often be tricky. This is where the powerful technique of partial fraction decomposition comes into play.

Integrals Ex 7.5 of your NCERT textbook focuses entirely on mastering this method. By breaking down complex rational functions into simpler fractions, we can transform an otherwise difficult integral into a sum of easily integrable terms. This page will guide you through the different cases of partial fractions, show you how to apply the method systematically, and provide worked examples and practice questions to ensure you thoroughly understand and can confidently solve problems from Integrals Ex 7.5. Let's make complex integrals simple!

Understanding Partial Fractions for Integration

Partial fraction decomposition is a technique used to integrate rational functions, which are functions expressed as a ratio of two polynomials, say P(x)/Q(x). While some simple rational functions can be integrated directly (like 1/x which integrates to log|x|), many complex ones require this special method. The core idea is to express the complex rational function P(x)/Q(x) as a sum of simpler rational functions (called partial fractions) that are much easier to integrate.

Before applying partial fractions, it's crucial to check one condition: the degree of the numerator P(x) must be strictly less than the degree of the denominator Q(x). If the degree of P(x) is greater than or equal to the degree of Q(x), you must first perform polynomial long division to express P(x)/Q(x) as a sum of a polynomial and a proper rational function (where the new numerator's degree is less than the denominator's). Once this condition is met, the process involves factoring the denominator Q(x) completely and then setting up the appropriate partial fraction form based on the nature of these factors. This systematic approach ensures that even intimidating-looking integrals become manageable.

Different Cases of Partial Fraction Decomposition

Case 1: Denominator has Non-repeated Linear Factors: If Q(x) can be factored into distinct linear factors like (x-a)(x-b)(x-c), then the rational function P(x)/Q(x) can be expressed as: A/(x-a) + B/(x-b) + C/(x-c). Here, A, B, C are constants that we need to find. We typically find these constants by substituting convenient values of x (the roots of the factors) or by comparing coefficients of powers of x on both sides.
Case 2: Denominator has Repeated Linear Factors: If Q(x) has a repeated linear factor, for example (x-a)^n, then for each such factor, the partial fraction expansion will include terms like: A_1/(x-a) + A_2/(x-a)^2 + ... + A_n/(x-a)^n. For instance, if Q(x) = (x-a)^2(x-b), the decomposition is A/(x-a) + B/(x-a)^2 + C/(x-b). Constants are found using substitution or by comparing coefficients.
Case 3: Denominator has Non-repeated Irreducible Quadratic Factors: An irreducible quadratic factor is a quadratic expression (ax^2+bx+c) that cannot be factored into linear factors with real coefficients (i.e., its discriminant b^2-4ac < 0). If Q(x) contains such a factor, say (x-a)(Ax^2+Bx+C), the corresponding partial fraction form is: A/(x-a) + (Bx+C)/(Ax^2+Bx+C). The numerator for the quadratic factor is a linear polynomial Bx+C.

Step-by-Step Integration Using Partial Fractions

Example 1: Non-repeated Linear Factors Integrate ∫(x / ((x+1)(x+2))) dx Step 1: Decompose the integrand into partial fractions. Let x / ((x+1)(x+2)) = A/(x+1) + B/(x+2) Multiply by (x+1)(x+2): x = A(x+2) + B(x+1) Step 2: Find the constants A and B. To find A, set x = -1: -1 = A(-1+2) + B(-1+1) => -1 = A(1) + B(0) => A = -1 To find B, set x = -2: -2 = A(-2+2) + B(-2+1) => -2 = A(0) + B(-1) => -2 = -B => B = 2 So, x / ((x+1)(x+2)) = -1/(x+1) + 2/(x+2) Step 3: Integrate the partial fractions. ∫(-1/(x+1) + 2/(x+2)) dx = ∫(-1/(x+1)) dx + ∫(2/(x+2)) dx = -log|x+1| + 2log|x+2| + C Final Answer: 2log|x+2| - log|x+1| + C or log| (x+2)^2 / (x+1) | + C
Example 2: Repeated Linear Factors Integrate ∫(3x+1 / ((x-1)^2(x+1))) dx Step 1: Decompose the integrand into partial fractions. Let (3x+1) / ((x-1)^2(x+1)) = A/(x-1) + B/(x-1)^2 + C/(x+1) Multiply by (x-1)^2(x+1): 3x+1 = A(x-1)(x+1) + B(x+1) + C(x-1)^2 Step 2: Find the constants A, B, and C. To find B, set x = 1: 3(1)+1 = A(0) + B(1+1) + C(0) => 4 = 2B => B = 2 To find C, set x = -1: 3(-1)+1 = A(0) + B(0) + C(-1-1)^2 => -2 = C(-2)^2 => -2 = 4C => C = -1/2 To find A, compare coefficients or substitute another value, e.g., x = 0: 1 = A(-1)(1) + B(1) + C(-1)^2 1 = -A + B + C Substitute B=2 and C=-1/2: 1 = -A + 2 - 1/2 1 = -A + 3/2 => A = 3/2 - 1 => A = 1/2 So, (3x+1) / ((x-1)^2(x+1)) = (1/2)/(x-1) + 2/(x-1)^2 - (1/2)/(x+1) Step 3: Integrate the partial fractions. ∫((1/2)/(x-1) + 2/(x-1)^2 - (1/2)/(x+1)) dx = (1/2)∫(1/(x-1)) dx + 2∫(1/(x-1)^2) dx - (1/2)∫(1/(x+1)) dx = (1/2)log|x-1| + 2 * (-1/(x-1)) - (1/2)log|x+1| + C Final Answer: (1/2)log|x-1| - 2/(x-1) - (1/2)log|x+1| + C

Exam Tips for Integrals by Partial Fractions

Mastering partial fractions requires attention to detail. Here are some critical tips for your CBSE exams:

Always Check Degrees First: Before anything else, compare the degree of the numerator P(x) with the degree of the denominator Q(x). If deg(P(x)) ≥ deg(Q(x)), perform polynomial long division. Failing to do this is a very common mistake and will lead to an incorrect partial fraction setup.
Factorize Denominator Completely: Ensure Q(x) is fully factored into its simplest linear or irreducible quadratic factors. Any error here will propagate throughout the entire solution.
Correct Partial Fraction Form: Carefully choose the correct partial fraction form based on the types of factors (non-repeated linear, repeated linear, irreducible quadratic). A mismatch here will make finding constants impossible or incorrect.
Efficiently Find Constants: While comparing coefficients always works, using the 'substitution method' (plugging in roots of the factors) can be much faster for linear factors. For irreducible quadratic factors, or when comparing coefficients, ensure you equate terms with the same powers of x accurately.
Don't Forget the Integration: After finding the constants and decomposing, remember to actually integrate each partial fraction. Common integrals you'll use are ∫(1/(ax+b)) dx = (1/a)log|ax+b| + C and ∫(1/(ax+b)^n) dx = (1/a) * (ax+b)^(1-n) / (1-n) + C for n ≠ 1.

Practice Questions with Solutions

Q: Integrate ∫(x / (x^2 - 9)) dx A: Step 1: Factorize the denominator: x^2 - 9 = (x-3)(x+3). Step 2: Decompose into partial fractions: x / ((x-3)(x+3)) = A/(x-3) + B/(x+3). Multiply by (x-3)(x+3): x = A(x+3) + B(x-3). Set x=3: 3 = A(6) => A = 1/2. Set x=-3: -3 = B(-6) => B = 1/2. So, x / ((x-3)(x+3)) = (1/2)/(x-3) + (1/2)/(x+3). Step 3: Integrate: ∫((1/2)/(x-3) + (1/2)/(x+3)) dx = (1/2)log|x-3| + (1/2)log|x+3| + C. Final answer: (1/2)log|(x-3)(x+3)| + C or (1/2)log|x^2-9| + C.
Q: Find ∫(3x - 2) / ((x+1)^2 + 1) dx (Correction: this is not a partial fraction question as denominator is irreducible quadratic, the original problem should be ∫(3x - 2) / ((x+1)^2(x+3)) dx for partial fractions with repeated linear factors. Let's use a simpler one: ∫(x^2 + 1) / (x^2 - 5x + 6) dx) A: Step 1: Check degrees. deg(P(x)) = 2, deg(Q(x)) = 2. Since degrees are equal, perform polynomial long division. (x^2 + 1) / (x^2 - 5x + 6) = 1 + (5x - 5) / (x^2 - 5x + 6). Step 2: Factorize the new denominator: x^2 - 5x + 6 = (x-2)(x-3). Step 3: Decompose the proper rational function: (5x - 5) / ((x-2)(x-3)) = A/(x-2) + B/(x-3). Multiply by (x-2)(x-3): 5x - 5 = A(x-3) + B(x-2). Set x=2: 5(2)-5 = A(2-3) => 5 = -A => A = -5. Set x=3: 5(3)-5 = B(3-2) => 10 = B => B = 10. So, (5x - 5) / ((x-2)(x-3)) = -5/(x-2) + 10/(x-3). Step 4: Integrate the original expression: ∫(1 + (-5/(x-2)) + (10/(x-3))) dx = ∫1 dx - 5∫(1/(x-2)) dx + 10∫(1/(x-3)) dx = x - 5log|x-2| + 10log|x-3| + C. Final answer: x - 5log|x-2| + 10log|x-3| + C.
Q: Evaluate ∫(1 / (x(x^2+1))) dx A: Step 1: Denominator has a non-repeated linear factor x and an irreducible quadratic factor x^2+1. Step 2: Decompose into partial fractions: 1 / (x(x^2+1)) = A/x + (Bx+C)/(x^2+1). Multiply by x(x^2+1): 1 = A(x^2+1) + (Bx+C)x. 1 = Ax^2 + A + Bx^2 + Cx. 1 = (A+B)x^2 + Cx + A. Step 3: Compare coefficients: Coeff of x^2: A+B = 0 Coeff of x: C = 0 Constant term: A = 1 From A=1 and A+B=0, we get 1+B=0 => B=-1. So, 1 / (x(x^2+1)) = 1/x - x/(x^2+1). Step 4: Integrate: ∫(1/x - x/(x^2+1)) dx = ∫(1/x) dx - ∫(x/(x^2+1)) dx For the second integral, let u = x^2+1, then du = 2x dx => x dx = (1/2)du. ∫(x/(x^2+1)) dx = ∫(1/2u) du = (1/2)log|u| = (1/2)log|x^2+1|. So, the integral is log|x| - (1/2)log|x^2+1| + C. Final answer: log|x| - (1/2)log(x^2+1) + C (since x^2+1 is always positive, absolute value is not needed).
Q: Find ∫(x^2) / ((x^2+4)(x^2+9)) dx A: Step 1: This is a special case. We can substitute y = x^2 to simplify the partial fraction decomposition for y / ((y+4)(y+9)). (Remember to substitute back before integration). y / ((y+4)(y+9)) = A/(y+4) + B/(y+9) y = A(y+9) + B(y+4) Set y=-4: -4 = A(-4+9) => -4 = 5A => A = -4/5. Set y=-9: -9 = B(-9+4) => -9 = -5B => B = 9/5. So, x^2 / ((x^2+4)(x^2+9)) = (-4/5)/(x^2+4) + (9/5)/(x^2+9). Step 2: Integrate the terms. ∫((-4/5)/(x^2+4) + (9/5)/(x^2+9)) dx = (-4/5)∫(1/(x^2+2^2)) dx + (9/5)∫(1/(x^2+3^2)) dx Use the formula ∫(1/(x^2+a^2)) dx = (1/a)arctan(x/a) + C. = (-4/5) (1/2)arctan(x/2) + (9/5) (1/3)arctan(x/3) + C = (-2/5)arctan(x/2) + (3/5)arctan(x/3) + C. Final answer: (3/5)arctan(x/3) - (2/5)arctan(x/2) + C.

Frequently Asked Questions

What is partial fraction decomposition?

Partial fraction decomposition is a technique used in integral calculus to express a complex rational function (a ratio of two polynomials) as a sum of simpler fractions. This transformation makes it much easier to integrate the original complex function.

When should I use partial fractions for integration?

You should use partial fractions when integrating rational functions `P(x)/Q(x)` where direct integration methods like substitution or standard formulas are not applicable. It is especially useful when the denominator `Q(x)` can be factored into linear or irreducible quadratic factors.

What if the degree of the numerator is greater than or equal to the denominator?

If the degree of the numerator `P(x)` is greater than or equal to the degree of the denominator `Q(x)`, you must first perform polynomial long division. This will reduce the rational function into a sum of a polynomial (which is easy to integrate) and a proper rational function, for which partial fraction decomposition can then be applied.

How do I find the constants (A, B, C) in partial fractions?

There are two primary methods: 1. **Substitution Method:** Substitute the roots of the linear factors (or other convenient `x` values) into the equation `P(x) = A(...) + B(...)`. This often quickly yields the values of constants for linear factors. 2. **Comparing Coefficients:** Expand the right-hand side of the partial fraction equation and group terms by powers of `x`. Then, equate the coefficients of corresponding powers of `x` on both sides of the equation to form a system of linear equations, which you can solve for the constants.