Integration of log x/1+x for x>0 then value of.(x)+f(1/x)

We need to evaluate the integral:

I = ∫ (log x) / (1 + x) dx

Step 1: Substituting x = 1/t
Let x = 1/t, so that dx = -dt/t². Substituting these in the given integral:

I = ∫ (log(1/t)) / (1 + 1/t) * (-dt/t²)

Since log(1/t) = -log t, we get:

I = ∫ (-log t) / ( (t + 1) / t ) * (-dt/t²)

Simplifying the denominator:

I = ∫ (-log t) * (t / (t + 1)) * (-dt/t²)

I = ∫ (log t) * (t / (t + 1)) * (dt/t²)

I = ∫ (log t) / (t + 1) dt

Step 2: Finding f(x) + f(1/x)
Let:
f(x) = ∫ (log x) / (1 + x) dx

From our substitution, we found that:
f(1/x) = ∫ (log (1/x)) / (1 + 1/x) dx

Since log(1/x) = -log x, we get:

f(1/x) = ∫ (-log x) / (1 + x) dx

Adding both expressions:

f(x) + f(1/x) = ∫ (log x) / (1 + x) dx + ∫ (-log x) / (1 + x) dx

The two integrals cancel out, giving:

f(x) + f(1/x) = 0