NDA Mathematics Question (2026) — Solution

Step-by-step solution

Let I = \,d (2+ )(3+4 ) Substitute = t, which gives - \,d = dt \,d = -dt I = -dt (2+t)(3+4t) Using partial fractions: -1 (t+2)(4t+3) = P t+2 + Q 4t+3 -1 = P(4t+3) + Q(t+2) Substituting t = -2, we get -1 = P(-8+3) P = 1 5 Substituting t = - 3 4 , we get -1 = Q (- 3 4 +2 ) Q = - 4 5 Therefore, the integral becomes: I = ( 1/5 t+2 - 4/5 4t+3 ) dt I = 1 5 |t+2| - 4 5 1 4 |4t+3| + C I = 1 5 |t+2| - 1 5 |4t+3| + C Substituting back t = : I = 1 5 |2+ | - 1 5 |3+4 | + C Comparing this with A |2+ | + B |3+4 |, we get A = 1 5 . Answer: 1 5