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Q10:
Integrate the rational functions in Exercises 1 to 21. 10. $\frac{2x-3}{(x^2-1)(2x+3)}$

Solution :

Step 1: Factorization and Decomposition Setup

The given integrand is $I = \int \frac{2x-3}{(x^2-1)(2x+3)} \, dx$. First, factorize the quadratic term $(x^2-1)$ as $(x-1)(x+1)$ [By difference of squares identity $a^2-b^2 = (a-b)(a+b)$].

The expression becomes: $\frac{2x-3}{(x-1)(x+1)(2x+3)}$.

We decompose this into partial fractions as follows:

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

Step 2: Determining Constants $A, B,$ and $C$

Multiplying both sides by the denominator $(x-1)(x+1)(2x+3)$, we obtain:

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

  • For $A$ (Let $x = 1$): $2(1)-3 = A(1+1)(2(1)+3) \Rightarrow -1 = A(2)(5) \Rightarrow A = -\frac{1}{10}$
  • For $B$ (Let $x = -1$): $2(-1)-3 = B(-1-1)(2(-1)+3) \Rightarrow -5 = B(-2)(1) \Rightarrow B = \frac{5}{2}$
  • For $C$ (Let $x = -\frac{3}{2}$): $2(-\frac{3}{2})-3 = C(-\frac{3}{2}-1)(-\frac{3}{2}+1) \Rightarrow -6 = C(-\frac{5}{2})(-\frac{1}{2}) \Rightarrow -6 = C(\frac{5}{4}) \Rightarrow C = -\frac{24}{5}$

Step 3: Integration of Partial Fractions

Substituting the constants back into the integral expression:

$I = \int \left( -\frac{1}{10(x-1)} + \frac{5}{2(x+1)} - \frac{24}{5(2x+3)} \right) dx$

Applying the linearity property of integrals $\int (af(x) + bg(x)) dx = a \int f(x) dx + b \int g(x) dx$ and the standard integral $\int \frac{1}{ax+b} dx = \frac{1}{a} \ln|ax+b| + C$:

$I = -\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{24}{5} \cdot \frac{1}{2} \ln|2x+3| + K$

$I = -\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{12}{5} \ln|2x+3| + K$

Step 4: Geometric Representation of the Integrand

x y f(x) = (2x-3)/((x^2-1)(2x+3))

Final Solution: $-\frac{1}{10} \ln|x-1| + \frac{5}{2} \ln|x+1| - \frac{12}{5} \ln|2x+3| + K$


More Questions from Class 12 Mathematics Integrals EXERCISE 7.5


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