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प्रश्न
\[\int\frac{2}{\left( e^x + e^{- x} \right)^2} dx\]
पर्याय
- \[\frac{- e^{- x}}{e^x + e^{- x}} + C\]
- \[- \frac{1}{e^x + e^{- x}} + C\]
- \[\frac{- 1}{\left( e^x + 1 \right)^2} + C\]
- \[\frac{1}{e^x - e^{- x}} + C\]
MCQ
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उत्तर
\[\frac{- e^{- x}}{e^x + e^{- x}} + C\]
\[\text{Let }I = \int\frac{2 dx}{\left( e^x + e^{- x} \right)^2}\]
\[ = \int\frac{2 dx}{\left( e^x + \frac{1}{e^x} \right)^2}\]
\[ = 2\int\frac{e^{2x} dx}{\left( e^{2x} + 1 \right)^2}\]
\[\text{Let }e^{2x} + 1 = t\]
\[ \Rightarrow e^{2x} \cdot 2 dx = dt\]
\[ \Rightarrow e^{2x} \cdot dx = \frac{dt}{2}\]
\[ \therefore I = 2 \times \frac{1}{2}\int\frac{dt}{t^2}\]
\[ = - \frac{1}{t} + C\]
\[ = - \frac{1}{e^{2x} + 1} + C ...............\left( \because t = e^{2x} + 1 \right)\]
Dividing numerator and denominator by ex
\[\Rightarrow I = \frac{- \frac{1}{e^x}}{e^x + \frac{1}{e^x}}\]
\[ = \frac{- e^{- x}}{e^x + e^{- x}} + C\]
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