find the antiderivative of the given derivative.\n\\( \\frac { d p } { d x } = \\frac { e ^ { x } + 2 e ^ {…

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

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

Answer

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\neq - 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)