What is the solution for f^'(x)=(1-e^x)/(9+e^x) ?

The solution for f^'(x)=(1-e^x)/(9+e^x) is f(x)= 1/9 x-10/9 ln(9+e^x)+c_{1}

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f^'(x)=(1-e^x)/(9+e^x)

f^{^{'}} (x) = \frac{1 - e ^{x}}{9 + e ^{x}}

f (x) = \frac{1}{9} x - \frac{10}{9} ln (9 + e^{x}) + c_{1}

Solution steps

f^{'} (x) = \frac{1 - e ^{x}}{9 + e ^{x}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

f (x) = \int \frac{1 - e ^{x}}{9 + e ^{x}} d x

\int \frac{1 - e ^{x}}{9 + e ^{x}} d x = \frac{1}{9} x - \frac{10}{9} ln (9 + e^{x}) + c_{1}

f (x) = \frac{1}{9} x - \frac{10}{9} ln (9 + e^{x}) + c_{1}

x y^{^{'}} + y = - 2 x y^{2}, y (1) = 2

\frac{\partial}{\partial y} (y^{2} sin (x y))

x \to 5 lim (\frac{2}{( x - 5 ) ^{6}})

x \to 1 lim (2 x^{- 1})

t e^{- 5 t} sin (5 t)

What is the solution for f^'(x)=(1-e^x)/(9+e^x) ?
The solution for f^'(x)=(1-e^x)/(9+e^x) is f(x)= 1/9 x-10/9 ln(9+e^x)+c_{1}