Question

( \int_0^1 \log(x + 1) , dx = )

• A. ( \log 2 - 1 )
• B. ( \log 2 + 1 )
• C. ( 2\log 2 + 1 )
• D. ( 2\log 2 - 1 )

Hint:

$\int \log x \, dx = x\log x - x$. Use this by shifting $x$ to $x+1$.

Answer 1

The Correct Option is D

Step 1: Integration by parts
(\int 1 \cdot \log(x+1) , dx = x\log(x+1) - \int \frac{x}{x+1} , dx).

Step 2: Simplify second term
(\int \frac{x+1-1}{x+1} , dx = \int (1 - \frac{1}{x+1}) , dx = x - \ln(x+1)).

Step 3: Combine and evaluate
([x\log(x+1) - x + \log(x+1)]_{0}^{1}).
(= [1\log 2 - 1 + \log 2] - [0 - 0 + 0] = 2\log 2 - 1).
Final Answer: (D)