First slide

Methods of integration

Question

$Evaluate \int \frac{e^{2 x} - 1}{2 x} d x$

Easy

A

$\log |e^{x} - e^{- x}| + C$
B

$\log |e^{x} + e^{- x}| + C$
C

$\frac{1}{2} \log |e^{x} + e^{- x}| + C$
D

None of these

Solution

$I = \int \frac{e^{2 x} - 1}{e^{2 x} + 1} d x = \int \frac{\frac{(e^{2 x} - 1)}{e^{x}}}{\frac{(e^{2 x} + 1)}{e^{x}}} d x = \int \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x$
$\begin{array}{l} \begin{array}{l} Let e^{x} + e^{- x} = t \\ or (e^{x} - e^{- x}) d x = d t \end{array} \\ ∴ I = \int \frac{d t}{t} \\ = \log | t | + C \\ = \log |e^{x} + e^{- x}| + C \end{array}$

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