For x > 0, if f(x) = ∫₁ˣ (log ₑt)/(1 + t)dt, then f(e) + f( 1/e ) is equal to :

Solution

Maths sir · 4 steps

Find f(1/x) by putting t = 1/u, then add it to f(x). The two integrands combine neatly.

1. Rewrite f(1/x). In f(1/x) = ∫₁^(1/x) (ln t)/(1+t) dt, put t = 1/u. Then dt = -du/u², ln t = -ln u and 1 + t = (1+u)/u. The limits 1 to 1/x become 1 to x.
2. Simplify. The two minus signs cancel, so f(1/x) = ∫₁ˣ (ln u)/u(1+u) du. The variable name does not matter, so call it t again.

   f(x) + f(1/x) = ∫₁ˣ ln t(1/(1+t) + 1/t(1+t))dt = ∫₁ˣ (ln t)/t dt

3. Integrate. The bracket is (t+1)/t(1+t) = 1/t. Since d/dt((ln t)²/2) = (ln t)/t, the sum equals (ln x)²/2 - 0 = (ln x)²/2.
4. So. At x = e, f(e) + f(1/e) = (ln e)²/2 = 1/2, which is option (a).

The idea to keep. When a question asks for f(x) + f(1/x), substitute t = 1/u in the second integral so both run from 1 to x, then add the integrands.

Answer from the NTA final answer key. Solution by PrepWiser.