Mathematics - Indefinite Integration Question with Solution | TestHub

We have, $R\left(x\right)=\int \frac{dx}{e^x+8e^{-x}+4e^{-3x}}-2\int \frac{dx}{e^{3x}+8e^x+4e^{-x}}$

$\Rightarrow R\left(x\right)=\int \frac{e^x\left(e^{2x}-2\right)}{e^{4x}+8e^{2x}+4}dx$

we get $R\left(t\right)=\int \frac{\left(t^2-2\right)dt}{t^4+8t^2+4}=\int \frac{\left(1-2t^{-2}\right)dt}{{\left(t+2t^{-1}\right)}^2+4}$

$=\frac{1}{2}{\tan }^{-1}\left(\frac{t+2t^{-1}}{2}\right)+K$

$\Rightarrow R\left(x\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{e^x+2e^{-x}}{2}\right)+K$

Hence, $\left(A,B,C\right)=\left({\tan }^{-1},e^x,2\right)$

$\therefore$ Option (b) is correct.