Question

The value of the integral \( \int_{1}^{e} \frac{1 + \log x}{3x} \, dx \) is equal to:

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{4} \)
• D. \( e \)
• E. \( \frac{1}{e} \)

Hint:

Always look for a function and its derivative within the integral. Here, seeing $1/x$ and $\log x$ together is a classic signal to use $u = \log x$ or $u = 1 + \log x$.

Answer 1

The Correct Option is B

Concept: We use the method of substitution. We notice that the derivative of \( \log x \) is \( \frac{1}{x} \), which is present in the integrand.

Step 1: Apply substitution.
Let \( u = 1 + \log x \). Then \( du = \frac{1}{x} \, dx \). Change the limits:
• When \( x = 1 \), \( u = 1 + \log 1 = 1 \).
• When \( x = e \), \( u = 1 + \log e = 1 + 1 = 2 \).

Step 2: Rewrite and evaluate the integral.
\[ \int_{1}^{e} \frac{1 + \log x}{3x} \, dx = \frac{1}{3} \int_{1}^{2} u \, du \] \[ = \frac{1}{3} \left[ \frac{u^2}{2} \right]_{1}^{2} = \frac{1}{6} [u^2]_{1}^{2} \]

Step 3: Final calculation.
\[ \frac{1}{6} (2^2 - 1^2) = \frac{1}{6} (4 - 1) = \frac{3}{6} = \frac{1}{2} \]