Question-167082

Question Number 167082 by cortano1 last updated on 06/Mar/22

Commented by cortano1 last updated on 06/Mar/22

\int ({fog}^{- 1}) (x) dx = x^{3} + c

\int g (x) f^{'} (x) dx = ?

Answered by greogoury55 last updated on 06/Mar/22

\int f (g^{- 1} (x)) d x = x^{3} + C

\Rightarrow f (g^{- 1} (x)) = 3 x^{2}

\Rightarrow f (g^{- 1} (g (x))) = 3 {(g (x))}^{2}

\Rightarrow f (x) = 3 {(g (x))}^{2}

\Rightarrow f^{'} (x) = 6 g (x) . g^{'} (x)

∴ \int g (x) f^{'} (x) d x = \int g (x) . 6 g (x) g^{'} (x) d x

= \int 6 {(g (x))}^{2} d (g (x))

= \frac{6}{3} {(g (x))}^{3} + C

= 2 {(g (x))}^{3} + C