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Question:

If ∫dx(x² - 2x + 10)²= A (tan⁻¹(x - 1/3) + f(x)x² - 2x + 10) + C where C is a constant of integration, then:

A=181andf(x)=3(x-1)

A=154andf(x)=9(x-1)²

A=154andf(x)=3(x-1)

A=127andf(x)=9(x-1)

Solution:

Correct option is D. A=154 and f(x)=3(x-1)
∫dx((x-1)²+9)² = ∫cos²θdθ (Put x-1=3tanθ) = ∫(1+cos2θ)/2 dθ = (θ/2 + sin2θ/4) + C
= (1/2)tan⁻¹((x-1)/3) + (1/4)sin(2tan⁻¹((x-1)/3)) + C
= (1/2)tan⁻¹((x-1)/3) + (1/2) * (2tan⁻¹((x-1)/3))/(1+((x-1)/3)²) + C
= (1/2)tan⁻¹((x-1)/3) + (1/2)*(3(x-1))/((x-1)²+9) + C
= (1/2)tan⁻¹((x-1)/3) + (3(x-1))/(2((x-1)²+9)) + C
Comparing with the given integral, A=154 and f(x) = 3(x-1)