Question

Evaluate the following : `int_0^1 (log(x + 1))/(x^2 + 1)*dx`

Answer 1

Let I = `int_0^1 (log(x + 1))/(x^2 + 1)*dx`

Put x = tan θ.
∴ dx = sec²θ·dθ 
and
x² + 1 = tan²θ + 1 = sec²θ

When x = 0, tan θ = 0    ∴ θ = 0

When x = 1, tan θ = 1    ∴ θ = `pi/(4)`

∴ I = `int_0^(pi/4) (log(tan theta + 1))/sec^2 theta* sec2 theta *d theta`

= `int_0^(pi/4) log(1 + tan theta)*d theta`           ...(1)

We use the property, `int_0^a f(x)*dx = int_0^a f(a - x)*dx`.

Here, a = ``pi/(4).

Hence changing θ by `pi/(4) - theta`, we have,

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

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

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

= `pi/(4) log 2 - "I"`

∴ 2I = `pi/(4) log 2`

∴ I = `pi/(8) log 2`.