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If $f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^n} - {x^{ - n}}}}{{{x^n} + {x^{ - n}}}},x > 1$ then $\int {\dfrac{{xf\left( x \right)\ln \left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right)}}{{\sqrt {\left( {1 + {x^2}} \right)} }}dx}$ isA) $\ln \left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right) - x + c$B) $\dfrac{1}{2}\left\{ {{x^2}\ln \left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right) - {x^2}} \right\} + c$C) $x\ln \left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right) - \ln \left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right) + c$D) None of these

Last updated date: 20th Sep 2024
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Hint: We have given a function $f\left( x \right)$ and we have to solve the integration. Firstly, we have to find the value of $f\left( x \right)$. The function $f\left( x \right)$ is $\dfrac{\infty }{\infty }$ form, so we cannot put the limit in it. We have to simplify it. Once we get the value of $f\left( x \right)$, we can solve the integration. In integration, firstly we will simplify the integral function by putting it equal to another variable. Then, we apply an integration formula to solve it.

Complete step-by-step solution:
We have given that $f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^n} - {x^{ - n}}}}{{{x^n} + {x^{ - n}}}},x > 1$ and
We have to calculate value of $\int {\dfrac{{xf\left( x \right)\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {\left( {1 + {x^2}} \right)} }}}$
Now $f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^n} - {x^{ - n}}}}{{{x^n} + {x^{ - n}}}},x > 0$
It is $\dfrac{\infty }{\infty }$ form as $x > 0$
$\Rightarrow \,\,f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^n} - \dfrac{1}{{{x^n}}}}}{{{x^n} + \dfrac{1}{{{x^n}}}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{{{x^{2n}} - 1}}{{{x^n}}}}}{{\dfrac{{{x^{2n}} + 1}}{{{x^n}}}}}$
The bottom of both numerator and denominator are hence cancel it, then
$\Rightarrow \,\,f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^{2n}} - 1}}{{{x^{2n}} + 1}}$
It is again $\dfrac{\infty }{\infty }$ form
Now, Take ${x^{2n}}$ as common in both numerator and denominator
$\Rightarrow \,\,f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{x^{2n}}\left( {1 - \dfrac{1}{{{x^{2n}}}}} \right)}}{{{x^{2n}}\left( {1 + \dfrac{1}{{{x^{2n}}}}} \right)}}$
$\Rightarrow \,\,\,f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {1 - \dfrac{1}{{{x^{2n}}}}} \right)}}{{\left( {1 + \dfrac{1}{{{x^{2n}}}}} \right)}} = \dfrac{{1 - \dfrac{1}{\infty }}}{{1 + \dfrac{1}{\infty }}} = \dfrac{{1 - 0}}{{1 + 0}} = 1$--------(1)
Now let consider
$\Rightarrow \,\,\,\,{\rm I} = \int {\dfrac{{x \cdot f\left( x \right)\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {\left( {1 + {x^2}} \right)} }}dx}$
From equation (1) the value of $f(x) = 1$, then
$\Rightarrow \,\,\,\,{\rm I} = \int {\dfrac{{x \cdot 1\ln \left( {x + \sqrt {1 + {x^2}} } \right)}}{{\sqrt {\left( {1 + {x^2}} \right)} }}dx}$------------(2)
Putting $x + \sqrt {1 + {x^2}} = t$-----------(a)
Differentiating both sides, we get
$\Rightarrow \,\,\,\left( {1 + \dfrac{1}{2}\dfrac{{2x}}{{\sqrt {1 + {x^2}} }}} \right)dx = dt$
On simplification, we get
$\Rightarrow \,\,\,\left( {1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}} \right)dx = dt$
$\Rightarrow \dfrac{{\sqrt {1 + {x^2}} + x}}{{\sqrt {1 + {x^2}} }}dx = dt$
$\Rightarrow$Now value of $\sqrt {1 + {x^2}} + x = t$ so, we have
$\Rightarrow \,\,\dfrac{t}{{\sqrt {1 + {x^2}} }}dx = dt$
$\Rightarrow \,\,\dfrac{{dx}}{{\sqrt {1 + {x^2}} }} = \dfrac{1}{t}dt$----------(b)
Again consider,
$\Rightarrow \,\,x + \sqrt {1 + {x^2}} = t$
Subtract x on both side, then
$\Rightarrow \,\,x + \sqrt {1 + {x^2}} - x = t - x$
$\Rightarrow \sqrt {1 + {x^2}} = t - x$
Squaring both sides, we get
$\Rightarrow 1 + {x^2} = {\left( {t - x} \right)^2}$
Apply the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ on RHS, then
$\Rightarrow 1 + {x^2} = {t^2} + {x^2} - 2tx$
Subtract ${x^2}$ on both side
$\Rightarrow 1 + {x^2} - {x^2} = {t^2} + {x^2} - 2tx - {x^2}$
On simplification, we get
$\Rightarrow \,\,\,1 = {t^2} - 2tx$
or
$\Rightarrow 2tx = {t^2} - 1$
Divide both side by 2t
$\Rightarrow x = \dfrac{{{t^2} - 1}}{{2t}}$---------(c)
Now substitute (a), (b) and (c) values in (2), then
$\Rightarrow \,\,\,I = \int {\dfrac{{{t^2} - 1}}{{2t}}\ln \left( t \right)\dfrac{{dt}}{t}}$
$\Rightarrow \,\,\,\,I = \dfrac{1}{2}\int {\dfrac{{{t^2} - 1}}{{{t^2}}}\ln \left( t \right)dt}$
$\Rightarrow \,\,\,I = \dfrac{1}{2}\int {\left( {1 - \dfrac{1}{{{t^2}}}} \right)\ln \left( t \right)dt}$
Apply integration separately
$\Rightarrow \,\,I = \dfrac{1}{2}\int {\ln \left( t \right)dt - \dfrac{1}{2}\int {\dfrac{{\ln \left( t \right)}}{{{t^2}}}dt} }$
In $\ln \left( t \right)$apply integration of product
$\Rightarrow \,\,\,I = \dfrac{1}{2}\left[ {t\left( {\ln \left( t \right) - 1} \right)} \right] - \dfrac{1}{2}\left[ { - \dfrac{1}{t}\ln t - \dfrac{1}{t}} \right] + c$
$\Rightarrow \,\,\,I = \dfrac{{t\ln \left( t \right)}}{2} - \dfrac{t}{2} + \dfrac{{\ln \left( t \right)}}{{2t}} + \dfrac{1}{{2t}} + c$
Substitute the value of t
${\rm I} = \dfrac{{x + \sqrt {{x^2} + 1} \ln \left( {x + \sqrt {{x^2} + 1} } \right)}}{2} + \dfrac{{x + \sqrt {{x^2} + 1} }}{2} + \dfrac{{\ln \left( {x + \sqrt {{x^2} + 1} } \right)}}{{2\left( {x + \sqrt {{x^2} + 1} } \right)}} + \dfrac{1}{{2\left( {x + \sqrt {{x^2} + 1} } \right)}} + C$

Hence the correct answer is option ‘D’.

Note: Integration is a way of adding slices to find the whole integration can be used to find are, volume and central points. It is used to find many useful quantities.
i) Limit of a function: The limit of a function is a fundamental concept in calculus and analysis concerning the behavior of the function near a particular input.
ii) Differentiation: The derivative of a function of a real variable measures the sensitivity to the change of a function with respect to change in argument.