Question

Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1)  "d"x`

Answer 1

Let I = `int_0^1 (log(x + 1))/(x^2 + 1)  "d"x`

Put x = tan θ

∴ dx = sec²θ dθ

When x = 0, θ = 0 and when x = 1, θ = `pi/4`

∴ I = `int_0^(pi/4)(log(tantheta + 1))/(tan^2theta + 1) xx sec^2theta  "d"theta`

= `int_0^(pi/4) (log(1 + tantheta))/(sec^2theta) xx sec^2theta  "d"theta`

∴ I = `int_0^(pi/4) log(1 + tan theta)  "d"theta`  ......(i)

= `int_0^(pi/4) log[1 + tan(pi/4 - theta)] "d"theta`  ......`[∵ int_0^"a" "f"(x)  "d"x = int_0^"a"  "f"("a" - x)  "d"x]`

= `int_0^(pi/4) log[1 + (tan  pi/4 - tantheta)/(1 + tan  pi/4 tan theta)] "d"theta`

= `int_0^(pi/4) log[1 + (1 - tan theta)/(1 + tan theta)]  "d"theta`

= `int_0^(pi/4) log[(1 + tan theta + 1 - tan theta)/(1 + tan theta)] "d"theta`

= `int_0^(pi/4) log[2/(1 + tan theta)] "d"theta`

= `int_0^(pi/4) [log2 - log(1 + tan theta)]  "d"theta`

= `log 2 int_0^(pi/4)  "d"theta - int_0^(pi/4) log(1 + tan theta)  "d"theta`

∴ I = `log 2[theta]_0^(pi/4) - "I"`     .....[From (i)]

∴ 2I = `log 2(pi/4 - 0)`

∴ I = `pi/8 log 2`