First slide

Evaluation of definite integrals

Question

If $\int \frac{e^{4 x} - 1}{e^{2 x}} \log (\frac{e^{2 x} + 1}{e^{2 x} - 1}) d x$
$= \frac{t^{2}}{2} \log t - \frac{t^{2}}{4} + \frac{u^{2}}{2} \log u - \frac{u^{2}}{4} + C$ then

Moderate

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

Solution

The integrand can be written as
$\begin{array}{l} (e^{2 x} - e^{- 2 x}) \log (e^{x} + e^{- x}) - (e^{2 x} - e^{- 2 x}) \log (e^{x} - e^{- x}) \\ = (e^{x} + e^{- x}) (e^{x} - e^{- x}) \log (e^{x} + e^{- x}) \\ - (e^{x} + e^{- x}) (e^{x} - e^{- x}) \log (e^{x} - e^{- x}) . \end{array}$
Hence the given integrals is equal to
$\begin{array}{l} \int t \log t d t - \int u \log u d u \\ (t = e^{x} + e^{- x}, u = e^{x} - e^{- x}) \\ = \frac{t^{2}}{2} \log t - \frac{1}{2} \int t d t + \frac{u^{2}}{2} \log u - \frac{1}{2} \int u d u \end{array}$
(integration by parts)
$= \frac{t^{2}}{2} \log t - \frac{t^{2}}{4} + \frac{u^{2}}{2} \log u - \frac{u^{2}}{4} + C$
where $t = e^{x} + e^{- x}$ and $u = e^{x} - e^{- x}$ .

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