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Q2:
Integrate the rational functions in Exercises 1 to 21. 2. $\frac{1}{x^2-9}$

Solution :

Given Variables & Initial Setup

We are required to evaluate the integral of the rational function:

$I = \int \frac{1}{x^2 - 9} \, dx$

This integral takes the form $\int \frac{1}{x^2 - a^2} \, dx$, where $a = 3$. We shall employ the Method of Partial Fractions to decompose the integrand.

Step 1: Factorizing the Denominator

Using the algebraic identity for the difference of squares, $a^2 - b^2 = (a - b)(a + b)$, we factor the denominator:

$x^2 - 9 = (x - 3)(x + 3)$ [By Algebraic Identity $a^2 - b^2$]

Step 2: Partial Fraction Decomposition

We express the integrand as a sum of partial fractions:

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

Multiplying both sides by $(x - 3)(x + 3)$ yields:

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

Step 3: Solving for Coefficients A and B

To find $A$ and $B$, we equate the coefficients or substitute convenient values for $x$:

  • Let $x = 3$: $1 = A(3 + 3) + B(0) \implies 6A = 1 \implies A = \frac{1}{6}$
  • Let $x = -3$: $1 = A(0) + B(-3 - 3) \implies -6B = 1 \implies B = -\frac{1}{6}$

Step 4: Integration

Substituting the values of $A$ and $B$ back into the integral:

$I = \int \left( \frac{1/6}{x - 3} - \frac{1/6}{x + 3} \right) \, dx$

$I = \frac{1}{6} \int \frac{1}{x - 3} \, dx - \frac{1}{6} \int \frac{1}{x + 3} \, dx$

Using the standard integral formula $\int \frac{1}{u} \, du = \ln|u| + C$ [By Integration Principle of Linear Forms]:

$I = \frac{1}{6} \ln|x - 3| - \frac{1}{6} \ln|x + 3| + C$

Step 5: Applying Logarithmic Properties

Using the property $\ln(m) - \ln(n) = \ln\left(\frac{m}{n}\right)$:

$I = \frac{1}{6} \ln\left| \frac{x - 3}{x + 3} \right| + C$

Integral of f(x) = 1/(x^2 - 9) Result: (1/6) * ln |(x-3)/(x+3)| + C

Final Solution: $\frac{1}{6} \ln\left| \frac{x - 3}{x + 3} \right| + C$


More Questions from Class 12 Mathematics Integrals EXERCISE 7.5


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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