Question

$calcule \: a \: integral \: \frac{5 - \times }{ { \times }^{2} + \times - 2 } dx$
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Answer 1

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Answer 2

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$\boxed{\begin{array}{l}\sf\dfrac{5-x}{x^2+x-2}=\dfrac{5-x}{(x-1)(x+2)}\\\underline{\rm escrevendo~como~soma~de~frac_{\!\!,}\tilde oes~parciais~temos\!:}\\\sf\dfrac{5-x}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}\\\\\sf A=\dfrac{5-x}{x+2}\bigg|_{x=1}=\dfrac{5-1}{2+1}=\dfrac{4}{3}\\\\\sf B=\dfrac{5-x}{x-1}\bigg|_{x=-2}=\dfrac{5-[-2]}{-2-1}=-\dfrac{7}{3}\\\\\sf\dfrac{5-x}{(x-1)(x+2)}=\dfrac{4}{3}\cdot\dfrac{1}{x-1}-\dfrac{7}{3}\cdot\dfrac{1}{x+2}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{5-x}{x^2+x-2}~dx=\int\dfrac{5-x}{(x-1)(x+2)}~dx\\\displaystyle\sf\dfrac{4}{3}\int\dfrac{dx}{x-1}-\dfrac{7}{3}\int\dfrac{dx}{x+2}\\\displaystyle\sf\int\dfrac{dx}{x-1}\\\underline{\rm fac_{\!\!,}a}\\\rm t=x-1\implies dt=dx\\\displaystyle\sf\int\dfrac{dx}{x-1}=\int\dfrac{dt}{t}=\ell n|t|+k\\\sf como~t=x-1~temos~\displaystyle\sf\int\dfrac{dx}{x-1}=\ell n|x-1|+k\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{dx}{x+2}\\\underline{\rm fac_{\!\!,}a}\\\sf s=x+2\implies ds=dx\\\displaystyle\sf\int\dfrac{dx}{x+2}=\int\dfrac{ds}{s}=\ell n|s|+k\\\sf como~s=x+2~temos~\displaystyle\sf\int\dfrac{dx}{x+2}=\ell n|x+2|+k\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{5-x}{(x-1)(x+2)}~dx=\dfrac{4}{3}\ell n |x-1|-\dfrac{7}{3}\ell n|x+2|+k\\\sf usando~as~propriedades\\\sf k\cdot\ell n (a)=\ell n(a)^k~e~\ell na-\ell nb=\ell n\bigg(\dfrac{a}{b}\bigg)~temos:\\\displaystyle\sf\int\dfrac{5-x}{(x-1)(x+2)}~dx=\ell n|x-1|^{\frac{4}{3}}-\ell n|x+2|^{\frac{7}{3}}+k\\\displaystyle\sf\int\dfrac{5-x}{x^2+x-2}~dx=\sf\ell n\bigg|\dfrac{(x-1)^{\frac{4}{3}}}{(x+2)^{\frac{7}{3}}}\bigg|+k\end{array}}$