Question

If \( F(x)=\int x(\log x)^{2} \, dx \) and \( F(e)=\frac{e^{2}}{4} \), then \( F(1) = \)

• A. \( 0 \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{3\log(2)}{4} \)

Hint:

Since \( \log 1 = 0 \), substituting \( x = 1 \) into your terms causes all components containing log to vanish completely, leaving only the pure polynomial constant fraction behind.

Answer 1

The Correct Option is B

Step 1: First integration by parts.
Take $u=(\log x)^2$ and $dv=x\,dx$, so $du=\frac{2\log x}{x}dx$ and $v=\frac{x^2}{2}$. \[ F(x)=\frac{x^2}{2}(\log x)^2-\int x\log x\,dx \]
Step 2: Second integration by parts.
For $\int x\log x\,dx$, take $u=\log x$ and $dv=x\,dx$: \[ \int x\log x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4} \]
Step 3: Combine the results.
\[ F(x)=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C \]
Step 4: Use $F(e)=\frac{e^2}{4}$.
Put $x=e$ (so $\log e=1$): \[ \frac{e^2}{4}=\frac{e^2}{2}-\frac{e^2}{2}+\frac{e^2}{4}+C\implies C=0 \]
Step 5: Substitute $x=1$.
Since $\log 1=0$, the first two terms vanish: \[ F(1)=0-0+\frac{1}{4} \]
Step 6: State the value.
\[ F(1)=\frac{1}{4} \] \[ \boxed{\tfrac{1}{4}} \]