If $f(x)=\lim _{n \rightarrow \infty} n^2\left(x^{\frac{1}{n}}-x^{\frac{1}{n+1}}\right), x>0$, then $\int x f(x) d x=$

• A

  $\frac{x^2}{2} \log x+C$
• B

  $\frac{x^2}{2} \log x+\frac{x^2}{4}+C$
• Correct answer
  C

  $\frac{x^2}{2} \log x-\frac{x^2}{4}+C$
• D

  $-\frac{x^2}{2} \log x+\frac{x^2}{4}+C$

Solution

Let $m=\frac{1}{n}$. So, as $n \rightarrow \infty \Rightarrow m \rightarrow 0$ $\begin{aligned} & \Rightarrow f(\mathrm{x})=\lim _{m \rightarrow 0} \frac{1}{m^2}\left(x^m-x^{\frac{m}{m+1}}\right) \\ & \Rightarrow \lim _{m \rightarrow 0} \frac{x^{\frac{m}{m+1}}}{m^2}\left[x^{\frac{m^2}{m+1}}-1\right] \\ & \Rightarrow \lim _{m \rightarrow 0} x^{\frac{m}{m+1}}\left[\frac{x^{\frac{m 2}{m+1}}-1}{\left(\frac{m^2}{m+1}\right)}\right] \cdot \frac{1}{(m+1)}=x^{\circ}[\log x] \cdot \frac{1}{(0+1)} \\ & =1 \cdot \log x \cdot 1 \Rightarrow f(x)=\log x\end{aligned}$ Now, $\int x f(x) d x=\int x \log x d x$ $=\log x \cdot \frac{x^2}{2}-\int \frac{x^2}{2} \cdot \frac{1}{x} d x+c=\frac{x^2}{2} \log x-\frac{x^2}{4}+c$

Hence, the correct answer is option: 'C'.