First slide

Methods of integration

Question

$\int \frac{1}{{(e^{x} + e^{- x})}^{2}} d x =$

Easy

A

$- \frac{1}{2 (e^{2 x} + 1)} + c$
B

$\frac{1}{2 (e^{2 x} + 1)} + c$
C

$- \frac{1}{e^{2 x} + 1} + c$
D

$None of these$

Solution

$\begin{array}{l} I = \int \frac{1}{{(e^{x} + e^{- x})}^{2}} d x = \int \frac{e^{2 x}}{{(e^{2 x} + 1)}^{2}} d x \\ Put e^{2 x} + 1 = t \Rightarrow 2 e^{2 x} d x = d t \\ \Rightarrow I = \frac{1}{2} \int \frac{1}{t^{2}} d t = - \frac{1}{2} \cdot \frac{1}{t} + c = - \frac{1}{2 (e^{2 x} + 1)} + c \end{array}$

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