Bonjour,
[tex]Soit \ f(x) = \frac{x^3+3 \sqrt{x}+1 }{x^2} \quad x>0 \\\\
\int f(x)dx = \int \frac{x^3+3 \sqrt{x}+1 }{x^2} dx \\\\
\int f(x)dx = \int [x+3x^{-3/2}+ \frac{1}{x^2} ]dx \\\\
\int f(x)dx = \int x \ dx +\int 3x^{-3/2} \ dx + \int \frac{1}{x^2} \ dx \\\\
\int f(x) dx =\boxed{ \frac{x^2}{2} - \frac{6}{ \sqrt{x} }- \frac{1}{x}+Constante} \\\\\\
Soit \ g(x) = cos(3x)\\
Une \ primitive \ de \ cos(u) \ est \ \frac{sin(u)}{u'} \\\\
\int g(x)dx = \int cos(3x)dx = \boxed{\frac{sin(3x)}{3} + Constante}[/tex]
