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$I = \int\limits_0^\infty {\frac{{{e^{ - nx}} - {e^{ - x}}}}{x}dx} = ?$

$n \in \mathbb{N}$

Let, ${e^{ - x}} = t$

$ \Rightarrow  - {e^{ - x}}dx = dt$

$ \Rightarrow  - tdx = dt$

$I = \int\limits_1^0 {\frac{{{t^n} - t}}{{ - \ln t}}\frac{{dt}}{{ - t}}}  =  - \int\limits_0^1 {\frac{{{t^{n - 1}} - 1}}{{\ln t}}dt} $

But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx = \ln (\lambda  + 1)} $ (Proof can be seen in this problem)

So, $I =  - \ln (n - 1 + 1) =  - \ln n$

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$