Question

$\displaystyle \sf \red{ \int^{\infty}_{0} \bigg(x - \frac{ {x}^{3} }{2} + ... \bigg)\bigg(1 + \frac{x {}^{2} }{4} + ... \bigg) \: dx}$​

Answer 1

$\\ \bigstar \large \: \sf \textcolor{blue}{ Given \: Expression :- }$

$\\ \quad \bullet \quad \sf \bigg(x - \dfrac{1}{x} \bigg) ^{ \dfrac{1}{2} } + \bigg(1 - \dfrac{1}{x} \bigg)^{ \dfrac{1}{2} }=x$

This expression can be written as follows :-

$\\ \quad \bullet \quad \underline{ \boxed{\sf \textcolor{Red}{\sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} + \sqrt{\bigg(1 - \dfrac{1}{x} \bigg)}=x} }} \frak{ \: - - - (i)}$

$\\ \\large \: \sf \textcolor{Blue}{ Solution :-}$

Multiply the Above eq (i) with the conjugates of LHS

$\\ \quad \longrightarrow \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} + \sqrt{\bigg(1 - \dfrac{1}{x} \bigg)} \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)} =x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \bigg[\sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} \bigg]^{2} - \bigg[\sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}\bigg]^{2} = x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \bigg(x - \dfrac{1}{x} \bigg)- \bigg( 1 - \dfrac{1}{x} \bigg) = x \times \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf \frac{x - 1}{x} = \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \sf 1 - \frac{1}{x} = \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)}$

$\\ \quad \longrightarrow \quad \underline{\boxed{ \sf \textcolor{red}{\sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} - \sqrt{ \bigg( 1 - \dfrac{1}{x} \bigg)} = 1 - \frac{1}{x} }}} \frak{ \: - - - (ii)}$

Now, by adding equations (i) and (ii) we get :-

$\\ \quad \longrightarrow \quad \underline{\boxed{ \sf \textcolor{red}{2\sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = x - \frac{1}{x} + 1}}} \frak{ \: - - - (iii)}$

$\\ \bigstar\large \: \sf \textcolor{Black}{ Assume \: The \: Following :-}$

$\quad \mapsto \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = a$

$\quad \therefore \quad \sf \bigg(x - \dfrac{1}{x} \bigg) = {a}^{2}$

Now, substituting the above values in equation (iii) , we get

$\\ \quad \longrightarrow \quad \sf 2a = {a}^{2} + 1$

$\\ \quad \longrightarrow \quad \sf {a}^{2} - 2a + 1 = 0$

$\quad \longrightarrow \quad \sf {(a - 1)}^{2} = 0$

$\\ \quad \longrightarrow \quad \boxed{\sf \textcolor{red}{a = \pm 1}}$

Now,

$\\ \quad : \implies \quad \sf \sqrt{ \bigg(x - \dfrac{1}{x} \bigg)} = a$

$\\ \quad : \implies \quad \sf \bigg(x - \dfrac{1}{x} \bigg) = 1$

$\\ \quad : \implies \quad \sf {x}^{2} - 1 = x$

$\\ \quad : \implies \quad \sf {x}^{2} - x - 1 = 0$

$\\ \quad : \implies \quad \sf x = \dfrac{ - ( - 1) \pm \sqrt{ { ( - 1)}^{2} - 4( - 1)(1)} }{2}$

$\\ \quad \therefore \quad \boxed{\frak{ \textcolor{Red}{ x = \dfrac{ 1 \pm \sqrt{5} }{2}}}} \bigstar$

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