For real numbers α, β, γ and δ, if
∫ ((x²-1)+tan⁻¹((x²+1)/x))/(x⁴+3x²+1)tan⁻¹((x²+1)/x)dx
=αlogₑ(tan⁻¹((x²+1)/x))+βtan⁻¹(γ(x²-1)/x)+δtan⁻¹((x²+1)/x)+C
where C is an arbitrary constant, then the value of 10(α + β γ + δ) is equal to ______________.

Solution

Maths sir · 5 steps

Split it in two: the first piece is T'/T, the second is a standard divide-by-x² integral.

1. Split the integrand. Let T = tan⁻¹((x²+1)/x). The integrand breaks into (x²-1)/((x⁴+3x²+1) T) + 1/(x⁴+3x²+1).
2. First piece. With u = x + 1/x, 1+u² = (x⁴+3x²+1)/x² and du/dx = (x²-1)/x². So dT/dx = (x²-1)/(x⁴+3x²+1), and the first piece is T'/T, which integrates to ln T. Hence α = 1.
3. Second piece. Divide top and bottom by x² and write 1/x² = 1/2[(1+1/x²) - (1-1/x²)]. The denominator x² + 3 + 1/x² equals (x-1/x)² + 5 and also (x+1/x)² + 1. Use the first form for the 1+1/x² part and the second form for the 1-1/x² part.

   ∫ dx/(x⁴+3x²+1) = 1/2√5 tan⁻¹((x²-1)/(√5 x)) - 1/2 tan⁻¹((x²+1)/x)

4. Read off constants. Comparing terms gives β = 1/2√5, γ = 1/√5 and δ = -1/2. So βγ = 1/10.
5. So. 10(α + βγ + δ) = 10(1 + 1/10 - 1/2) = 10 × 0.6 = 6.

The common slip: 16. Taking δ = +1/2 instead of -1/2 gives 10(1 + 0.1 + 0.5) = 16. The 1-1/x² part carries a minus sign.

The idea to keep. For ∫ dx/(x⁴ + kx² + 1), divide by x² and split 1/x² into 1/2[(1+1/x²) - (1-1/x²)]. Then substitute x ∓ 1/x.

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