Question

Calcule a integral por frações parciais.

SOS​

Answer 1

Resposta:

OLÁ

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• >>>>> INTEGRAL<<<<<

$\sf \displaystyle\int \dfrac{dx}{x^2+x-2}~dx$

$\sf \displaystyle\int \dfrac{1}{\left(x+\dfrac{1}{2} \right)^2 -\dfrac{4}{9}}$

$\sf \displaystyle\int \dfrac{4}{4u^2-9}}du$

$\sf 4\times \displaystyle\int \dfrac{1}{4u^2-9}}du$

$\sf 4\times \displaystyle\int\dfrac{1}{6(v^2-1)}}dv$

$\sf 4\times \dfrac{1}{6}\times\displaystyle\int\dfrac{1}{v^2-1} dv$

$\sf 4\times \dfrac{1}{6}\times\left(-\displaystyle\int\dfrac{1}{-v^2+1} dv\right)$

$\sf 4\times\dfrac{1}{6}\left(\left-\left(\dfrac{in|\dfrac{2}{3}\left(x+\dfrac{1}{2}\right)+1| }{2} -\dfrac{in|\dfrac{2}{3}\left(x+\dfrac{1}{2}\right) }{2} \right)\right):-\dfrac{1}{3}(in|\dfrac{4}{3}+ \dfrac{2}{3})$

$\boxed{\bold{\displaystyle{\clubsuit\ \spadesuit\ \maltese\ \sf \red{-\frac{1}{3}\left(in|\frac{4}{3}+\frac{2}{3} x|-in| -\frac{2}{3}+\frac{2}{3}x|\right)}+C }}}\ \checkmark$← RESPOSTA.

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