If f(x)=limn→∞ xn−x−nxn+x−n,x>1 then ∫xf(x)log⁡x+1+x21+x2dx is

f′(x)=limn→∞ xn−x−nxn+x−n=limn→∞ 1−(1/x)2n1+(1/x)2n=1(1/x<1) Using integration by parts , we get I=∫xlog⁡x+1+x21+x2dx=1+x2log⁡x+1+x2dx−∫dx=1+x2log⁡x+1+x2−x+C.

If f(x)=limn→∞ xn−x−nxn+x−n,x>1 then ∫xf(x)log⁡x+1+x21+x2dx is

f′(x)=limn→∞ xn−x−nxn+x−n=limn→∞ 1−(1/x)2n1+(1/x)2n=1(1/x<1) Using integration by parts , we get I=∫xlog⁡x+1+x21+x2dx=1+x2log⁡x+1+x2dx−∫dx=1+x2log⁡x+1+x2−x+C.

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If $f (x) = lim_{n \to \infty} \frac{x^{n} - x^{- n}}{x^{n} + x^{- n}}, x > 1$ then
$\int \frac{x f (x) \log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}$ dx is

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$(x - 1) \log (x + \sqrt{1 + x^{2}}) + C$

none of these

$\log (x + \sqrt{1 + x^{2}}) - x + C$

$(\frac{1}{2}) x^{2} (\log (x + \sqrt{1 + x^{2}}) - 1) + C$

answer is D.

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Detailed Solution

$f^{'} (x) = lim_{n \to \infty} \frac{x^{n} - x^{- n}}{x^{n} + x^{- n}} = lim_{n \to \infty} \frac{1 - (1 / x)^{2 n}}{1 + (1 / x)^{2 n}} = 1$

$(1 / x < 1)$

Using integration by parts , we get

$\begin{array}{l} I = \int \frac{x \log (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}} d x \\ = \sqrt{1 + x^{2}} \log (x + \sqrt{1 + x^{2}}) d x - \int d x \\ = \sqrt{1 + x^{2}} \log (x + \sqrt{1 + x^{2}}) - x + C . \end{array}$