How do you express $\frac{1}{e^{x} (e^{x} + 1)}$ $1/(e^x(e^x+1))$ in partial fractions?

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How do you express $\frac{1}{e^{x} (e^{x} + 1)}$ $1/(e^x(e^x+1))$ in partial fractions?

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Answer 1 · 2016-06-11T14:01:34

$\frac{1}{e^{x} (e^{x} + 1)} = \frac{1}{e^{x}} - \frac{1}{e^{x} + 1}$ $1/(e^x(e^x+1)) = 1/e^x-1/(e^x+1)$

Explanation:

Making $\frac{1}{e^{x} (e^{x} + 1)} = \frac{A}{e^{x}} + \frac{B}{e^{x} + 1}$ $1/(e^x(e^x+1)) = A/e^x+B/(e^x+1)$ after equating numerator coefficients we have

${ (A+B=0), (A=1) :}$

Solving for $A,B$ we obtain

$1/(e^x(e^x+1)) = 1/e^x-1/(e^x+1)$

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How do you express $\frac{1}{e^{x} (e^{x} + 1)}$ $1/(e^x(e^x+1))$ in partial fractions?

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