What is the solution for derivative of f^{-1}(x)=2x+ln(x+1) ?

The solution for derivative of f^{-1}(x)=2x+ln(x+1) is f(x)= x/(x^2+xln(1+x)+ln(1+x)-x-1+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of f^{-1}(x)=2x+ln(x+1)

\frac{d}{d x} f^{- 1} (x) = 2 x + ln (x + 1)

f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

Solution steps

\frac{d}{d x} (f (x)^{- 1} (x)) = 2 x + ln (x + 1)

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

then

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

f (x)^{- 1} x = \int 2 x + ln (x + 1) d x

\int 2 x + ln (x + 1) d x = x^{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c_{1}

f (x)^{- 1} x = x^{2} + x ln (x + 1) + ln (x + 1) - x - 1 + c_{1}

Isolate

f (x) : f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

f (x) = \frac{x}{x ^{2} + x ln ( 1 + x ) + ln ( 1 + x ) - x - 1 + c _{1}}

\int (1 - x)^{- 2} d x

\frac{\partial ^{2}}{\partial x \partial y} (x y^{2} - 6 x^{2} - 3 y^{2})

(7.4) (3.8)

x^{2} + 6 x - 1

\int - 2 x^{4} tan (x^{5}) d x

What is the solution for derivative of f^{-1}(x)=2x+ln(x+1) ?
The solution for derivative of f^{-1}(x)=2x+ln(x+1) is f(x)= x/(x^2+xln(1+x)+ln(1+x)-x-1+c_{1)}