Let f(x)=∫ √x/(1+x)²dx(x≥ 0). Then f(3) – f(1) is eqaul to :

Solution

Maths sir · 4 steps

A √x on top and 1+x below: put x = tan² θ and both roots and powers vanish.

1. Definite integral. Since f' (x) = √x/(1+x)², the difference f(3) - f(1) is ∫₁³ √x/(1+x)² dx. The constant of integration cancels.
2. Substitute. Put x = tan² θ, so √x = tan θ, 1 + x = sec² θ and dx = 2 tan θ sec² θ dθ. The limits become θ = π/4 at x = 1 and θ = π/3 at x = 3.

   ∫_(π/4)^(π/3) (tan θ · 2 tan θ sec² θ)/(sec⁴ θ) dθ = ∫_(π/4)^(π/3) 2 sin² θ dθ = [θ - (sin 2θ)/2]_(π/4)^(π/3)

3. Put in limits. At π/3: π/3 - √3/4. At π/4: π/4 - 1/2. Subtracting gives π/12 + 1/2 - √3/4.
4. So. f(3) - f(1) = π/12 + 1/2 - √3/4, which is option (b).

The idea to keep. When you see √x together with 1 + x, try x = tan² θ. Then 1 + x = sec² θ and the integrand usually becomes a plain power of sin θ or cos θ.

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