What is expand-KL(U(t-(e^{xt}-1)/x)) ?

The solution to expand-KL(U(t-(e^{xt}-1)/x)) is -LKUt+(Le^{tx}KU)/x-(LKU)/x

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expand-KL(U(t-(e^{xt}-1)/x))

expand

- K L (U (t - \frac{e ^{x t} - 1}{x}))

- L K U t + \frac{L e ^{t x} K U}{x} - \frac{L K U}{x}

Solution steps

- K L (U (t - \frac{e ^{x t} - 1}{x}))

Remove parentheses:

(a) = a

= - K LU (t - \frac{e ^{x t} - 1}{x})

= - L K U (t - \frac{e ^{t x} - 1}{x})

Apply the distributive law:

a (b - c) = ab - a c

a = - K LU, b = t, c = \frac{e ^{x t} - 1}{x}

= - K LU t - (- K LU) \frac{e ^{x t} - 1}{x}

Apply minus-plus rules

- (- a) = a

= - L K U t + L \frac{e ^{t x} - 1}{x} K U

L \frac{e ^{t x} - 1}{x} K U = \frac{L e ^{t x} K U - L K U}{x}

= - L K U t + \frac{L e ^{t x} K U - L K U}{x}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{L e ^{t x} K U - L K U}{x} = \frac{L e ^{t x} K U}{x} - \frac{L K U}{x}

= - L K U t + \frac{L e ^{t x} K U}{x} - \frac{L K U}{x}

\frac{( x ^{2} + 2 x )}{( x + 1 ) ^{2}}

(ln (3 x) - 5)^{2}

\frac{9}{( 6 + x ) ^{3}}

\frac{( 3 x - 2 ) ^{4}}{4}

(1 - cos (x))^{4}

What is expand-KL(U(t-(e^{xt}-1)/x)) ?
The solution to expand-KL(U(t-(e^{xt}-1)/x)) is -LKUt+(Le^{tx}KU)/x-(LKU)/x