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Sagot :

Réponse:

G(x)= 3RACINE carré de(x^2+1)

Réponse :

G(x) = [tex]3\sqrt{x^2+1}+C[/tex]

H(x) = [tex]-\frac{1}{x^2+3}+C[/tex]

J(x) = [tex]\frac{1}{3}e^{x^3}+C[/tex]

Explications étape par étape :

G(x) = [tex]\int \frac{3x}{\sqrt{x^2+1}}dx[/tex] = [tex]3\cdot \int \frac{x}{\sqrt{x^2+1}}dx[/tex] = [tex]3\cdot \frac{1}{2}\cdot \int \frac{1}{\sqrt{u}}du[/tex] = [tex]3\cdot \frac{1}{2}\cdot \int \:u^{-\frac{1}{2}}du[/tex] = [tex]3\cdot \frac{1}{2}\cdot \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}[/tex] = [tex]3\cdot \frac{1}{2}\cdot \frac{\left(x^2+1\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}[/tex] => [tex]3\sqrt{x^2+1}+C[/tex]

H(x) =  [tex]\int \frac{2x}{\left(x^2+3\right)^2}dx[/tex] = [tex]2\cdot \int \frac{x}{\left(x^2+3\right)^2}dx[/tex] = [tex]2\cdot \int \frac{1}{2u^2}du[/tex] = [tex]2\cdot \frac{1}{2}\cdot \int \:u^{-2}du[/tex] = [tex]2\cdot \frac{1}{2}\cdot \frac{u^{-2+1}}{-2+1}[/tex] = [tex]2\cdot \frac{1}{2}\cdot \frac{\left(x^2+3\right)^{-2+1}}{-2+1}[/tex] => [tex]-\frac{1}{x^2+3}+C[/tex]

J(x) = [tex]\int \:x^2e^{x^3}dx[/tex] = [tex]\int \frac{e^u}{3}du[/tex] = [tex]\frac{1}{3}\cdot \int \:e^udu[/tex] = [tex]\frac{1}{3}e^u[/tex] =>  [tex]\frac{1}{3}e^{x^3}+C[/tex]  

** [tex]dx = \frac{1}{3x^2}du[/tex]

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