What is expand (ln(1+1/x)-1/(x+1))(1+1/x)^x ?

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

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expand (ln(1+1/x)-1/(x+1))(1+1/x)^x

expand

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

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

Solution steps

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

Apply the distributive law:

a (b - c) = ab - a c

a = (1 + \frac{1}{x})^{x}, b = ln (1 + \frac{1}{x}), c = \frac{1}{x + 1}

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

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

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

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

\frac{d ^{2}}{d x ^{2}} ((y) ((3 x^{2} - 1)^{2}))

\frac{( 0.675 + 2 x ) ^{2}}{( 0.5643 - x ) ( 0.7653 - x )}

(x - 2) (x - 3) (x - 5) (x + 5)

(x + 2 - 3) (x + 2 + 3)

(2 - cos (x))^{2}

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