Question

If $\int \frac{e^{4x}-1}{e^{2x}}\log \left(\frac{e^{2x}+1}{e^{2x}-1}\right)ck=\frac{t^2}{2}\log t-\frac{t^2}{4}-\frac{u^2}{2}\log u+\frac{u^2}{4}+c$ then

• A. $t=e^{-x}-e^x,u=e^x+e^{-x}$
• B. $t=e^x-e^{-x},u=e^x+e^{-x}$
• C. $t=e^x+e^{-x},u=e^x-e^{-x}$
• D. $u=e^{-x}-e^x,t=e^x+e^{-x}$

Answer 1

The correct option is C $t=e^x+e^{-x},u=e^x-e^{-x}$

$\int \frac{e^{4x}-1}{e^{2x}}\log \left(\frac{e^{2x}+1}{e^{2x}-1}\right)dx$
$=\left(\log \left(\frac{e^{2x}+1}{e^{2x}-1}\right)(\frac{e^{2x}}{2}+\frac{e^{-2x}}{2})\right)-\int (\frac{e^{2x}}{2}+\frac{e^{-2x}}{2})\frac{\frac{d}{dx}\left(\frac{e^{2x}+1}{e^{2x}-1}\right)}{\frac{e^{2x}+1}{e^{2x}-1}}dx$
$=\log \left(\frac{e^{2x}+1}{e^{2x}-1}\right)\left(\frac{e^{2x}-e^{-2x}}{2}\right)-\int (\frac{e^{2x}}{2}+\frac{e^{-2x}}{2})\left(\frac{-e^{2x}(e^{2x}+1)}{(e^{2x}-1)}\right)dx$
$=\frac{t^2}{2}\log t-\frac{t^4}{4}-\frac{u^2}{2}\log u+\frac{u^2}{4}+c$
Where $t=e^x+e^{-x}$ and $u=e^x-e^{-x}$