Calculer une intégrale d'une composition de fonctions usuelles en passant par la primitive directement Exercice

\int_{0}^{1} (x+1)^2 \, \mathrm{d}x = \dfrac{7}{3}

\int_{0}^{1} (x+1)^2 \, \mathrm{d}x = \dfrac{5}{3}

\int_{0}^{1} (x+1)^2 \, \mathrm{d}x = \dfrac{4}{3}

\int_{0}^{1} (x+1)^2 \, \mathrm{d}x = \dfrac{1}{3}

\int_{0}^{1} \dfrac{1}{(x+1)^2} \, \mathrm{d}x = \dfrac{1}{2}

\int_{0}^{1} \dfrac{1}{(x+1)^2} \, \mathrm{d}x = -\dfrac{1}{2}

\int_{0}^{1} \dfrac{1}{(x+1)^2} \, \mathrm{d}x = \dfrac{3}{2}

\int_{0}^{1} \dfrac{1}{(x+1)^2} \, \mathrm{d}x = \dfrac{5}{2}

\int_{0}^{1} \dfrac{1}{(3x-4)^2} \, \mathrm{d}x = \dfrac{1}{4}

\int_{0}^{1} \dfrac{1}{(3x-4)^2} \, \mathrm{d}x = -\dfrac{1}{4}

\int_{0}^{1} \dfrac{1}{(3x-4)^2} \, \mathrm{d}x = \dfrac{5}{12}

\int_{0}^{1} \dfrac{1}{(3x-4)^2} \, \mathrm{d}x = -\dfrac{5}{12}

\int_{0}^{1} \dfrac{1}{\sqrt{2x+1}} \, \mathrm{d}x = \sqrt{3} - 1

\int_{0}^{1} \dfrac{1}{\sqrt{2x+1}} \, \mathrm{d}x = \sqrt{3}

\int_{0}^{1} \dfrac{1}{\sqrt{2x+1}} \, \mathrm{d}x = \sqrt{3} - \sqrt{2}

\int_{0}^{1} \dfrac{1}{\sqrt{2x+1}} \, \mathrm{d}x = \sqrt{2}

\int_{0}^{1} (2x-1)^2 \, \mathrm{d}x = \dfrac{1}{3}

\int_{0}^{1} (2x-1)^2 \, \mathrm{d}x = \dfrac{4}{3}

\int_{0}^{1} (2x-1)^2 \, \mathrm{d}x = \dfrac{2}{3}

\int_{0}^{1} (2x-1)^2 \, \mathrm{d}x = \dfrac{5}{3}