If f(α)=∫₁^α (log ₁₀ t)/(1+t) dt, α>0, then f(e³)+f(e⁻³) is equal to :

Solution

Maths sir · 4 steps

Put t = 1/u in f(1/α) and the two integrands add up to (log₁₀ t)/t.

1. Flip the second integral. Since e⁻³ = 1/e³, look at f(1/α) = ∫₁^(1/α) (log₁₀ t)/(1+t) dt. Put t = 1/u, so dt = -du/u² and the limits become 1 to α. Also log₁₀ t = -log₁₀ u and 1 + t = (u+1)/u.
2. Simplify. The two minus signs cancel, which gives f(1/α) = ∫₁^α (log₁₀ u)/u(1+u) du. It now has the same limits as f(α), so the two integrals can be added under one sign.

   f(α)+f(1/α) = ∫₁^α log₁₀ t(1/(1+t)+1/t(1+t))dt = ∫₁^α (log₁₀ t)/t dt

3. Integrate. Write log₁₀ t = (ln t)/(ln 10). Since ∫ (ln t)/t dt = (ln t)²/2, the sum equals (ln α)²/(2 ln 10).
4. So. With α = e³, ln α = 3, so f(e³) + f(e⁻³) = 9/(2 logₑ 10). That is option (d).

The idea to keep. When a question asks for f(α)+f(1/α), substitute t = 1/u in one integral. Bring both to the same limits and add: the two integrands usually combine into something easy.

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