MHTCET 22 Apr 2025 Shift 1 Paper

Concept:Use partial fractions to break the integrand, then apply the standard integral formula: $$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log\left|\frac{x-a}{x+a}\right| + c$$Explanation:Let the integrand be split as:$$\frac{2x^2+3}{(x^2-1)(x^2-4)} = \frac{A}{x^2-4} + \frac{B}{x^2-1}$$Multiply both sides by $(x^2-1)(x^2-4)$:$$2x^2 + 3 = A(x^2-1) + B(x^2-4)$$Comparing coefficients of $x^2$ and constant term:$$A + B = 2 \quad \text{and} \quad -A - 4B = 3$$Solving these equations gives:$$A = \frac{11}{3}, \quad B = \frac{-5}{3}$$Now integrate term by term:$$\int \frac{2x^2+3}{(x^2-1)(x^2-4)} dx = \int \frac{11/3}{x^2-4} dx - \int \frac{5/3}{x^2-1} dx$$Using the standard formula for $x^2 - 4$ (where $a=2$) and $x^2 - 1$ (where $a=1$):$$= \frac{11}{3} \cdot \frac{1}{4} \log\left|\frac{x-2}{x+2}\right| - \frac{5}{3} \cdot \frac{1}{2} \log\left|\frac{x-1}{x+1}\right| + c$$$$= \frac{11}{12} \log\left|\frac{x-2}{x+2}\right| + \frac{5}{6} \log\left|\frac{x+1}{x-1}\right| + c$$Combine using logarithm properties:$$= \log\left[\left(\frac{x-2}{x+2}\right)^{\frac{11}{12}} \cdot \left(\frac{x+1}{x-1}\right)^{\frac{5}{6}}\right] + c$$Comparing with the given form, we get:$$a = \frac{11}{12} \quad \text{and} \quad b = \frac{5}{6}$$Therefore:$$a + b = \frac{11}{12} + \frac{5}{6} = \frac{11}{12} + \frac{10}{12} = \frac{21}{12}$$Answer:$$a + b = \frac{21}{12}$$Hence, the correct option is B. $\frac{21}{12}$.