tex]Answer the given question....

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October 2022 1 44 Report

[tex]\footnotesize\displaystyle \sf \int \bigg( \sum_{n = 1} ^{ \infty } \bigg( \frac{d}{dx} \bigg( \lim_{h \to0} \bigg( \frac{ \frac{1}{(x + {h}^{2} )} - \frac{1}{ {x}^{2} } }{h} \bigg) \bigg) \bigg)^{n} \bigg) \: [/tex]

Answer the given question.

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EngrJohann

Verified answer

SOLUTION:

First, let's simplify the integral.

$\footnotesize\displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left( \dfrac{d}{dx} \left( \lim_{h \to 0} \left( \dfrac{ \dfrac{1}{(x + h)^{2} } - \dfrac{1}{x^2 } }{h} \right) \right) \right)^{n} \right)\ dx$

Notice that the derivative of a function, say $\sf f(x) = \dfrac{1}{x^2}$ , is expressed as a limit in the integrand. To show it, we have

$= \footnotesize \displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left( \dfrac{d}{dx} \left( \lim_{h \to 0} \left( \dfrac{ \dfrac{x^2 - (x + h)^2}{x^2(x + h)^{2} }}{h} \right) \right) \right)^{n} \right)\ dx$

Simplifying,

$= \footnotesize \displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left( \dfrac{d}{dx} \left( \lim_{h \to 0} \left(\dfrac{-2xh - h^2}{x^2(x + h)^{2} } \cdot \dfrac{1}{h} \right) \right) \right)^{n} \right)\ dx$

$= \footnotesize \displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left( \dfrac{d}{dx} \left( \lim_{h \to 0} \left(\dfrac{-2x - h}{x^2(x + h)^{2} }\right) \right) \right)^{n} \right)\ dx$

$= \footnotesize \displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left( \dfrac{d}{dx} \left(-\dfrac{2}{x^3} \right) \right)^{n} \right)\ dx$

$= \footnotesize \displaystyle \sf \int \left( \sum_{n = 1}^{ \infty } \left(\dfrac{6}{x^{4}}\right)^n \right)\ dx$

By infinite geometric series,

$= \footnotesize \displaystyle \sf \int \left( \dfrac{\dfrac{6}{x^4}}{1 - \dfrac{6}{x^4}}\right)\ dx\ for\ |x| > \sqrt[4]{6}$

Now we are left with the integral below which can be solved by partial fraction decomposition.

$\implies \displaystyle \sf \int \dfrac{6}{x^4 - 6}\ dx$

$\footnotesize \displaystyle \sf = \dfrac{6}{2\sqrt{6}} \int \left(\dfrac{1}{x^2 - \sqrt{6}} - \dfrac{1}{x^2 + \sqrt{6}} \right)\ dx$

$\sf \footnotesize = \dfrac{3}{\sqrt{6}} \left(\dfrac{1}{2\sqrt[4]{6}} \ln\Bigg | \dfrac{x - \sqrt[4]{6}}{x + \sqrt[4]{6}}\Bigg| - \dfrac{1}{\sqrt[4]{6}}\arctan\left(\dfrac{x}{\sqrt[4]{6}}\right) \right) + C$ for $\sf |x| > \sqrt[4]{6}$

ANSWER:

$\boxed{ \footnotesize \begin{array}{c} \sf \dfrac{3}{\sqrt{6}} \left(\dfrac{1}{2\sqrt[4]{6}} \ln\Bigg | \dfrac{x - \sqrt[4]{6}}{x + \sqrt[4]{6}}\Bigg| - \dfrac{1}{\sqrt[4]{6}}\arctan\left(\dfrac{x}{\sqrt[4]{6}}\right) \right) + C \\ \\ \sf for\ |x| > \sqrt[4]{6} \end{array}}$