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: Understanding the Question:
Evaluate a definite integral of a logarithmic function using integration by parts.
Step 3: Detailed Explanation:
Let \( I = \int_0^1 \log(x+1) \cdot 1 \, dx \).
Using parts: \( u = \log(x+1), dv = dx \implies v = x \).
\[ \int \log(x+1) \, dx = x \log(x+1) - \int \frac{x}{x+1} \, dx \]
\[ = x \log(x+1) - \int \left( 1 - \frac{1}{x+1} \right) dx \]
\[ = x \log(x+1) - [x - \log(x+1)] = (x+1)\log(x+1) - x \]
Applying limits from 0 to 1:
Upper limit (\( x=1 \)): \( (1+1)\log(2) - 1 = 2\log 2 - 1 \)
Lower limit (\( x=0 \)): \( (0+1)\log(1) - 0 = 0 \)
Result: \( 2\log 2 - 1 \).
Step 4: Final Answer:
The value is \( 2\log 2 - 1 \).