Question

If $\int_{}^{}\frac{e^{4x} - 1}{e^{2x}}$log $\left( \frac{e^{2x} + 1}{e^{2x} - 1} \right)$ dx = $\frac{t^{2}}{2}$log t – $\frac{t^{2}}{4}$

i + $\frac{u^{2}}{2}$ 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. None of these

Answer 1

Answer: (C)

t = e^x + e^–x, u = e^x – e^–x

Answer 2

The integrand can be written as

(e^2x – e^–2x) log (e^x + e^–x) – (e^2x – e^–2x) log (e^x – e^–x)

= (e^x + e^–x) (e^x –e^–x) log (e^x + e^–x) – (e^x + e^–x) (e^x – e^–x)

log (e^x – e^–x).

Hence the given integral is equal to

–$\int_{}^{}{u\log udu}$ (t = e^x + e^–x, u = e^x – e^–x)

= log t – $\frac { 1 } { 2 }$ $\int_{}^{}t$dt + $\frac{u^{2}}{2}$ log u – $\frac{1}{2}$ $\int_{}^{}u$ du

(integration by parts)

= log t –$\frac{t^{2}}{4}$ + $\frac{u^{2}}{2}$ log u $\frac { \mathrm { u } ^ { 2 } } { 4 }$ – + c,

where t = e^x + e^–x and u = e^x – e^–x.