Question

\[ \int \left( \frac{\log_e t}{1+t} + \frac{\log_e t}{t(1+t)} \right) dt \]

• A. \(\frac{(\log_e t)^2}{2} + C\)
• B. \(\frac{t^2 (\log_e t)^2}{2} + C\)
• C. \(\frac{(1+\log_e t)^2}{2} + C\)
• D. \(\frac{(\log_e t)^2}{2t^2} + C\)
• E. \(\frac{(\log_e t)^2}{2} + \frac{1}{(1+t)^2} + C\)

Hint:

In integrals involving logarithmic functions, combining terms and using integration by parts are effective strategies for simplification.

Answer 1

The Correct Option is A

First, simplify the integrand by combining the terms: \[ \frac{\log_e t}{1+t} + \frac{\log_e t}{t(1+t)} = \frac{\log_e t (1 + \frac{1}{t})}{1+t} = \frac{\log_e t (1 + t^{-1})}{1+t} \] \[ = \frac{\log_e t (t+1) t^{-1}}{1+t} = \frac{\log_e t}{t} \] Now, integrate the simplified expression: \[ \int \frac{\log_e t}{t} \, dt \] Using the integration by parts formula, let: \( u = \log_e t \) and \( dv = \frac{1}{t} dt \). Then, \( du = \frac{1}{t} dt \) and \( v = \log_e t \). 
Apply integration by parts: \[ \int u \, dv = uv - \int v \, du \] \[ = (\log_e t)(\log_e t) - \int (\log_e t) \frac{1}{t} dt \] \[ = (\log_e t)^2 - \int \frac{\log_e t}{t} dt \] Let \( I = \int \frac{\log_e t}{t} dt \), then: \[ I = (\log_e t)^2 - I \] \[ 2I = (\log_e t)^2 \] \[ I = \frac{(\log_e t)^2}{2} \] Thus, the integral evaluates to: \[ \int \frac{\log_e t}{t} dt = \frac{(\log_e t)^2}{2} + C \]