Question

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

Answer 1

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

Use substitution x = `1/t` idea via symmetry

` x = 1/t, dx = -1/t^2 dt`

When x = 0 ⇒ t = ∞,

when x = 1 ⇒ t = 1

`I = int_1^∞ ln(1+1/t)/(1+t^2) dx`

Now add the two forms of I:

`2I = int_0^1 (ln(1+x) + ln(1+1/x))/(1+x^2) dx`

Simplify logarithm

`ln(1+x) + ln (1+1/x)`

`= ln ((1+x)((1+x)/x))`

`= ln ((1+x)^2/x)`

`2I = int_0^1 (ln((1+x)^2/x))/(1+x^2) dx`

`2I = int_0^1 (2 ln (1+x) - ln x)/(1+x^2) dx`

`2I = 2I - int_0^1 lnx/(1+x^2) dx`

`int_0^1 ln x/(1+x^2) dx = 0`

`int_0^1 lnx/(1+x^2) dx = -pi/4 ln2`

`2I = 2I + pi/4 ln2 => I = pi/8 ln2`

`int_0^1 ln(1+x)/(1+x^2) dx = pi/8 ln2`