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Question-204262




Question Number 204262 by DEGWE last updated on 10/Feb/24
Answered by Frix last updated on 10/Feb/24
Charles−Ange LAISANT (1905):  ∫f^(−1) (x)dx=xf^(−1) (x)−(F○f^(−1) )(x)+C  with F(x)=∫f(x)dx
$$\mathrm{Charles}−\mathrm{Ange}\:\mathrm{LAISANT}\:\left(\mathrm{1905}\right): \\ $$$$\int{f}^{−\mathrm{1}} \left({x}\right){dx}={xf}^{−\mathrm{1}} \left({x}\right)−\left({F}\circ{f}^{−\mathrm{1}} \right)\left({x}\right)+{C} \\ $$$$\mathrm{with}\:{F}\left({x}\right)=\int{f}\left({x}\right){dx} \\ $$

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