MHT CET202521 Apr 2025Morning ShiftMathematicsIndefinite IntegrationActual
x^2-6 x-16 ~d x equals.
Options
- A( x-3 2 ) x^2-6 x-16 + 5 2 (x-3+ x^2-6 x-16 )+ c where c is the constant of integration
- B( x-3 2 ) x^2-6 x-16 - 25 2 (x-3+ x^2-6 x-16 )+c where c is the constant of integration
- C( x-3 2 ) x^2-6 x-16 + 25 2 (x-3+ x^2-6 x-16 )+ c where c is the constant of integration
- D( x-3 2 ) x^2-6 x-16 - 5 2 (x-3+ x^2-6 x-16 )+c , where c is the constant of integration
Correct answer
B. ( x-3 2 ) x^2-6 x-16 - 25 2 (x-3+ x^2-6 x-16 )+c where c is the constant of integration
Step-by-step solution
Evaluation of the Integral To evaluate x^2-6x-16 d x , complete the square for the quadratic expression: x^2 - 6x - 16 = (x-3)^2 - 25 = (x-3)^2 - 5^2 . The integral becomes (x-3)^2 - 5^2 d x , which matches the standard form u^2 - a^2 d u with u = x-3 and a = 5 . Using the standard formula: u^2 - a^2 d u = u 2 u^2 - a^2 - a^2 2 |u + u^2 - a^2 | + C . Substitute u = x-3 and a = 5 : = x-3 2 (x-3)^2 - 5^2 - 25 2 |(x-3) + (x-3)^2 - 5^2 | + C . Simplify to: = x-3 2 x^2 - 6x - 16 - 25 2 |x-3 + x^2 - 6x - 16 | + C . This