Relax

Respuesta :

konrad509
[tex]\displaystyle \int \dfrac{4x-8}{x^2-4x-3}\, dx=\\ 2\int \dfrac{2x-4}{x^2-4x-3}\, dx=(*)\\\\\\ \int \dfrac{f'(x)}{f(x)}=\ln(f(x))+C\\\\\\ (*)=2\ln(x^2-4x-3)+C[/tex]

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