Question

Evaluate:

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

Answer 1

We know that

`(d(tan^-1 x))/dx = 1/(1 + x^2)`

Since we have `1/(1 + x^2)`, we put x = tan θ

Thus, dx = sec² θ dθ

Change the limits of integration:

When x = 0, tan θ = 0 ⇒ θ = 0

When x = 1, tan θ = 1 ⇒ θ = `π/4`

The denominator becomes:

1 + x²

= 1 + tan² θ

= sec² θ

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

After substitution:

`I = int_0^(π/4) (log(1 + tanθ))/(1 + tan^2θ) xx sec^2θ  dθ`

= `int_0^(π/4) log (1 + tanθ)dθ`  ...(1)

Now, we know that

`int_0^a f(x) dx = int_0^a f(a - x) dx`

Applying this property to equation 1, with a = `π/4` and the variable being θ:

= `int_0^(π/4) log [1 + tan(π/4 - θ)]dθ`

Using tan (A − B) = `(tan A - tan B)/(1 + tanA tanB)`:

= `int_0^(π/4) log[1 + (tan(π/4) - tan θ)/(1 + tan(π/4) tan θ)] dθ`

= `int_0^(π/4) log [1 + (1 - tan θ)/(1 + tan θ)]dθ`

= `int_0^(π/4) log [(1 + tanθ + 1 - tanθ)/(1 + tanθ)]dθ`

= `int_0^(π/4) log [2/(1 + tanθ)]dθ`

Using the logarithm property `log (a/b)` = log a − log b

= `int_0^(π/4) [log 2 - log(1 + tan θ)] dθ`  ...(2)

Now, adding (1) and (2):

I + I = `int_0^(π/4) log (1 + tanθ)dθ + int_0^(π/4) [log 2 - log(1 + tan θ)] dθ`

2I = `int_0^(π/4) [log(1 + tan θ) + log 2 - log(1 + tanθ)] dθ`

2I = `int_0^(π/4) log 2 dθ`

Since log 2 is a constant:

2I = `log 2 int_0^(π/4) 1 dθ`

2I = `log 2 xx |θ|_0^(π/4)`

2I = `log 2 xx (π/4 - 0)`

2I = `π/4 log 2`

I = `π/8 log 2`