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

解

\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}}

解答ステップ

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

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

であれば

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}

分離する

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}}