Question

\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]

Options

• \[\frac{- e^{- x}}{e^x + e^{- x}} + C\]
• \[- \frac{1}{e^x + e^{- x}} + C\]
• \[\frac{- 1}{\left( e^x + 1 \right)^2} + C\]
• \[\frac{1}{e^x - e^{- x}} + C\]

Answer 1

\[\frac{- e^{- x}}{e^x + e^{- x}} + C\]

 

 

\[\text{Let }I = \int\frac{2 dx}{\left( e^x + e^{- x} \right)^2}\]

\[ = \int\frac{2 dx}{\left( e^x + \frac{1}{e^x} \right)^2}\]

\[ = 2\int\frac{e^{2x} dx}{\left( e^{2x} + 1 \right)^2}\]

\[\text{Let }e^{2x} + 1 = t\]

\[ \Rightarrow e^{2x} \cdot 2 dx = dt\]

\[ \Rightarrow e^{2x} \cdot dx = \frac{dt}{2}\]

\[ \therefore I = 2 \times \frac{1}{2}\int\frac{dt}{t^2}\]

\[ = - \frac{1}{t} + C\]

\[ = - \frac{1}{e^{2x} + 1} + C ...............\left( \because t = e^{2x} + 1 \right)\]

Dividing numerator and denominator by e^x

\[\Rightarrow I = \frac{- \frac{1}{e^x}}{e^x + \frac{1}{e^x}}\]

\[ = \frac{- e^{- x}}{e^x + e^{- x}} + C\]