Question

\(\int \frac{logx}{(1+logx)^2}dx =\)

• A. \(\frac{1}{1+logx}+c\)
• B. \(-\frac{1}{1+logx}+c\)
• C. \(\frac{x}{1+logx}+c\)
• D. \(-\frac{x}{1+logx}+c\)

Hint:

Write log x over (1+log x) squared as a derivative of x over (1+log x).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Write the numerator as \((1+\log x)-1\) so that the integrand splits into two simpler parts.

Step 2: Split
\[ \frac{\log x}{(1+\log x)^2}=\frac1{1+\log x}-\frac1{(1+\log x)^2} \]

Step 3: Integrate the first term by parts
\[ \int\frac{1}{1+\log x}dx=\frac{x}{1+\log x}+\int\frac{x\cdot\frac1x}{(1+\log x)^2}dx=\frac{x}{1+\log x}+\int\frac{1}{(1+\log x)^2}dx \]

Step 4: Combine
The last integral cancels the second term in the split:
\[ \int\frac{\log x}{(1+\log x)^2}dx=\frac{x}{1+\log x}+c \]

Step 5: Check
Differentiating \(\frac{x}{1+\log x}\) gives \(\frac{(1+\log x)-1}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}\). So option (C).

Final Answer:
The antiderivative is x divided by (1 + log x), option (C). \[ \boxed{\frac{x}{1+\log x}+c} \]