If ∫ dx/(x²+x+1)²=atan⁻¹((2x+1)/√3)+b((2x+1)/(x²+x+1))+C, x > 0 where C is the constant of integration, then the value of 9(√3a+b) is equal to _____________.

Solution

Maths sir · 4 steps

Complete the square, then a tan substitution turns this integral into one of cos² θ.

1. Complete the square. x² + x + 1 = (x + 1/2)² + 3/4. Put x + 1/2 = √3/2 tan θ, so dx = √3/2 sec² θ dθ and x² + x + 1 = 3/4 sec² θ.

   ∫ (√3/2 sec² θ dθ)/(9/16 sec⁴ θ) = 8√3/9∫ cos² θ dθ = 4√3/9 θ + 2√3/9 sin 2θ + C

2. Back to x. tan θ = (2x+1)/√3, so θ = tan⁻¹((2x+1)/√3). Also 1 + tan² θ = 4(x²+x+1)/3, which gives sin 2θ = (2 tan θ)/(1+tan² θ) = √3/2 × (2x+1)/(x²+x+1).
3. Find a and b. a = 4√3/9 and b = 2√3/9 × √3/2 = 1/3. So √3a + b = 4/3 + 1/3 = 5/3.
4. So. 9(√3a + b) = 9 × 5/3 = 15.

The idea to keep. For 1/(quadratic)², complete the square and use a tan substitution. sec⁴ θ in the denominator becomes cos² θ, and sin 2θ turns back into a rational term.

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