If f(x)=limn→∞ n2x1/n−x1/(n+1),x>0 then ∫xf(x)dx is equal to

f(x)=limn→∞ n2x1/(n+1)x1n−1n+1−1=limn→∞ x1/(n+1)x1n(n+1)−1n(n+1)n2×1n(n+1)=log⁡x ∫xfxdx=∫logx.xdx =logx x22-∫1x.x22dx =logx x22-x24+c

If f(x)=limn→∞ n2x1/n−x1/(n+1),x>0 then ∫xf(x)dx is equal to

f(x)=limn→∞ n2x1/(n+1)x1n−1n+1−1=limn→∞ x1/(n+1)x1n(n+1)−1n(n+1)n2×1n(n+1)=log⁡x ∫xfxdx=∫logx.xdx =logx x22-∫1x.x22dx =logx x22-x24+c

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If $f (x) = lim_{n \to \infty} n^{2} (x^{1 / n} - x^{1 / (n + 1)}), x > 0$ then $\int xf (x) dx$ is equal to

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$x^{2} \log x - \frac{1}{2} x^{2} + c$

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

$\frac{x^{2}}{2} + c$

answer is D.

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Detailed Solution

$\begin{array}{l} \begin{array}{l} f (x) = lim_{n \to \infty} n^{2} x^{1 / (n + 1)} [x^{\frac{1}{n} - \frac{1}{n + 1}} - 1] \\ = lim_{n \to \infty} \frac{x^{1 / (n + 1)} [x^{\frac{1}{n (n + 1)}} - 1]}{\frac{n (n + 1)}{n^{2}} \times \frac{1}{n (n + 1)}} \\ = \log x \end{array} \int x f (x) dx = \int \log x . x d x = \log x \frac{x^{2}}{2} - \int \frac{1}{x} . \frac{x^{2}}{2} d x = \log x \frac{x^{2}}{2} - \frac{x^{2}}{4} + c \end{array}$