What is expand-1/2 (1+e^{2x}-ln|1+e^{2x}|) ?

The solution to expand-1/2 (1+e^{2x}-ln|1+e^{2x}|) is -1/2-(e^{2x})/2+(ln|1+e^{2x}|)/2

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expand-1/2 (1+e^{2x}-ln|1+e^{2x}|)

expand

- \frac{1}{2} (1 + e^{2 x} - ln 1 + e^{2 x})

- \frac{1}{2} - \frac{e ^{2 x}}{2} + \frac{ln 1 + e ^{2 x}}{2}

Solution steps

- \frac{1}{2} (1 + e^{2 x} - ln 1 + e^{2 x})

Distribute parentheses

= (- \frac{1}{2}) \cdot 1 + (- \frac{1}{2}) e^{2 x} + (- \frac{1}{2}) (- ln 1 + e^{2 x})

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 1 \cdot \frac{1}{2} - \frac{1}{2} e^{2 x} + \frac{1}{2} ln 1 + e^{2 x}

Simplify

- 1 \cdot \frac{1}{2} - \frac{1}{2} e^{2 x} + \frac{1}{2} ln 1 + e^{2 x} : - \frac{1}{2} - \frac{e ^{2 x}}{2} + \frac{ln 1 + e ^{2 x}}{2}

= - \frac{1}{2} - \frac{e ^{2 x}}{2} + \frac{ln 1 + e ^{2 x}}{2}

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What is expand-1/2 (1+e^{2x}-ln|1+e^{2x}|) ?
The solution to expand-1/2 (1+e^{2x}-ln|1+e^{2x}|) is -1/2-(e^{2x})/2+(ln|1+e^{2x}|)/2