Question

∫3z+1/(z^2-4)^2
Como posso resolver essa integral urgente! !!
Não tenho mais informações foi assim que o professor deu no trabalho so a integral sem enunciado

Answer 1

Vamos utilizar a seguinte propriedade:

$\Large\boxed{\boxed{\mathsf{\displaystyle\int(C)~dt=c~\cdot~t}}}}}}}}}}}$

Veja:

$\mathsf{\displaystyle\int\dfrac{3z+1}{(z^{2}-4)^{2}}}}}}}\\\\\\\\\\ \mathsf{\dfrac{3z+1}{(z^{2}-4)^{2}}~\cdot~\mathsf{x}+C}}}}\\\\\\\\\ \mathsf{\dfrac{3zx+x}{(z^{2}-4)^{2}}}}+\mathsf{C}}}}}}}}}\\\\\\\\\ \mathsf{\dfrac{3zx+x}{(z^{2})^{2}+2z^{2}~\cdot~(-4)+(-4)^{2}}}}}}}+\mathsf{C}}}\\\\\\\\\\ \mathsf{\dfrac{3zx+x}{z^{2~\cdot~2}+16+(-8)~z^{2}}}}+\mathsf{C}}}}\\\\\\\\\\\ \large\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\mathsf{\displaystyle\int\dfrac{3z+1}{(z^{2}-4)^{2}}=\dfrac{3zx+x}{z^{4}-8z^{2}+16}}+\mathsf{C}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

Espero que te ajude. '-'