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)