Q.10) Definite integral

Q : Evaluate ∫[0 to 1] ln (x + 1)/(x^2 + 1) dx

SOLUTION :

Answer ► I = (π/8) * ln 2


Consider I = ∫[0 to 1] ln(x + 1)/(x^2 + 1) dx
Let x = tan θ, then dx = sec² θ dθ
For limits; when x = 0, θ = 0 and when x = 1,θ = π/4

therefore, I = ∫[0 to 1] ln (x + 1)/(x^2 + 1) dx
=> ∫[0 to π/4] ln (tan θ + 1) sec² θ dθ/(tan² θ + 1)
=> ∫[0 to π/4] ln (tan θ + 1) sec² θ dθ / sec² θ

=> ∫[0 to π/4] ln (1 + tan θ) dθ - - - - - (i)
I = ∫[0 to π/4] ln [1 + tan (π/4 - θ) ] dθ
=> ∫[0 to π/4] ln [1 + { tan (π/4) - tan θ}/ {1 + tan (π/4) * tan θ} ] dθ
=> ∫[0 to π/4] ln [1 + {1 - tan θ}/ {1 + tan θ} ] dθ
=> ∫[0 to π/4] ln [{1 + tan θ + 1 - tan θ}/ {1 + tan θ} ] dθ
=> ∫[0 to π/4] ln [ 2 / {1 + tan θ} ] dθ

=> ∫[0 to π/4] [ln 2 - ln (1 + tan θ) ] dθ - - - - - - (ii)

Adding (i) & (ii),
2I = ∫[0 to π/4] [ ln 2 ] dθ
=> [0 to π/4] [θ * ln 2 ]
=> π/4 * ln 2

therefore, I = π/8 * ln 2 ► Answer

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