If ∫ (x²+1)eˣ/(x+1)²dx=f(x)eˣ+C, where C is a constant, then d³f/dx³ at x = 1 is equal to :

Solution

Maths sir · 3 steps

Split the integrand into f + f' and the integral writes itself as eˣ f(x).

1. ∫ eˣ(f(x) + f'(x))dx = eˣf(x) + C

2. Find the pair. Take f(x) = (x-1)/(x+1) = 1 - 2/(x+1), so f'(x) = 2/(x+1)². Then f + f' = ((x-1)(x+1) + 2)/(x+1)² = (x²+1)/(x+1)², which is exactly the given integrand.
3. Differentiate again. From f'(x) = 2(x+1)⁻² we get f''(x) = -4(x+1)⁻³ and f'''(x) = 12(x+1)⁻⁴.
4. So. At x = 1, f'''(1) = 12/2⁴ = 12/16 = 3/4, which is option (b).

The idea to keep. When you see eˣ times a messy fraction, try to split the fraction as f + f'. A square in the denominator usually means f has that bracket to the first power.

Answer from the NTA final answer key. Solution by PrepWiser.