Question

find the antiderivative of the given derivative.
\\( \frac { d p } { d x } = \frac { e ^ { x } + 2 e ^ { - 2 x } } { ( e ^ { x } - e ^ { - 2 x } ) ^ { 3 } } \\)
\\( p = \square \\)

Explanation:

Step1: Substitute \( u = e^{x}-e^{-2x} \)

Let \( u = e^{x}-e^{-2x} \), then \( du=(e^{x}+2e^{-2x})dx \)

Step2: Integrate \( \frac{du}{u^{3}} \)

The integral \( \int\frac{e^{x}+2e^{-2x}}{(e^{x}-e^{-2x})^{3}}dx=\int\frac{du}{u^{3}} \)
Using the power - rule for integration \( \int u^{n}du=\frac{u^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), here \(n=-3\)
\(\int u^{-3}du=\frac{u^{-3 + 1}}{-3+1}+C=\frac{u^{-2}}{-2}+C=-\frac{1}{2u^{2}}+C\)

Step3: Substitute back \( u = e^{x}-e^{-2x} \)

Substitute \(u = e^{x}-e^{-2x}\) into the result, we get \(p=-\frac{1}{2(e^{x}-e^{-2x})^{2}}+C\)

Answer:

\(p =-\frac{1}{2(e^{x}-e^{-2x})^{2}}+C\)