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: Approach
Verify the answer by differentiation.

Step 2: Differentiate option (C)
Let $F=\dfrac{x}{1+L}$ where $L=\log x$. By the quotient rule:
\[ F'=\frac{(1+L)\cdot1-x\cdot\frac1x}{(1+L)^2}=\frac{L}{(1+L)^2} \]

Step 3: Compare
This is exactly the integrand, so $F$ is an antiderivative.

Step 4: Other options
Option (A) and (B) are missing the factor $x$, and option (D) has the wrong sign, so their derivatives do not equal the integrand. Hence (C).

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