Question

Ajutați-ma va rog cu rezolvarea integrala, pas cu pas, a exercițiului din foto.
Va mulțumesc! ​

Answer 1

Răspuns:

Explicație pas cu pas:

Voi efectua pe operatii...

$1.~1+\frac{2x+2}{x^{2}-3x-4} =1+\frac{2(x+1)}{x^{2}-4x+x-4}=1+\frac{2(x+1)}{x(x-4)+(x-4)} =1+\frac{2(x+1)}{(x-4)(x+1)}=1+\frac{2}{x-4} =\frac{x-4}{x-4} + \frac{2}{x-4} =\frac{x-4+2}{x-4} =\frac{x-2}{x-4} \\2.~\frac{x-2}{x-4} :\frac{x(x+2)}{x-4}= \frac{x-2}{x-4} *\frac{x-4}{x(x+2)}= \frac{x-2}{x(x+2)}\\ 3.~ \frac{x-2}{x(x+2)}-\frac{x+2}{x(x-2)} =\frac{(x-2)^{2}-(x+2)^{2}}{x(x-2)(x+2)}=\frac{((x-2)-(x+2))((x-2)+(x+2))}{x(x-2)(x+2)} =\frac{-4(2x)}{x(x-2)(x+2)} =\frac{-8}{(x-2)(x+2)}$

$4.~\frac{-8}{(x-2)(x+2)} *\frac{2-x}{2}= \frac{-8}{(x-2)(x+2)}*\frac{-(x-2)}{2}=\frac{4}{x+2}\\Deci~E(x)=\frac{4}{x+2}$