Thank you for reporting, we will resolve it shortly
Q.
Let $e$ be the eccentricity of hyperbola and $f ( e )$ be the eccentricity of its conjugate hyperbola then $\int\limits_1^{\sqrt{2}}(f(e)+f(f(e))) d e$ is equal to
Conic Sections
Solution:
Let eccentricity of conjugate hyperbola be e'
$\therefore \frac{1}{ e ^2}+\frac{1}{ e ^{\prime 2}}=1 \Rightarrow \frac{1}{ e ^{\prime 2}}=1-\frac{1}{ e ^2} \Rightarrow e ^{\prime}=\frac{ e }{\sqrt{ e ^2-1}} $
$\therefore f ( e )=\frac{ e }{\sqrt{ e ^2-1}}$
and $f(f(e))=\frac{f(e)}{\sqrt{f^2(e)-1}}=\frac{\frac{e}{\sqrt{e^2-1}}}{\sqrt{\frac{e^2}{e^2-1}-1}}=e$
$\therefore$ Given integral $=\int\limits_1^{\sqrt{2}}\left(\frac{ e }{\sqrt{ e ^2-1}}+ e \right) de =\left(\sqrt{ e ^2-1}+\frac{ e ^2}{2}\right)_1^{\sqrt{2}}=(1+1)-\left(0+\frac{1}{2}\right)=\frac{3}{2}$