Question

Por favor ajuda a resolver está Integral​

Answer 1

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$\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx\\\sf x^2+3x+2\\\sf s=-\dfrac{b}{a}=-\dfrac{3}{1}=-3\\\sf p=\dfrac{c}{a}=\dfrac{2}{1}=2\\\sf dois~n\acute umeros~que~somados~d\tilde ao-3~e~multiplicados~d\tilde ao~2\\\sf s\tilde ao~-2~e~-1~pois~(-2)+(-1)=-3~e~(-2)\cdot(-1)=2\\\sf logo~x_1=-1~e~x_2=-2\\\sf portanto~ax^2+bx+c=a\cdot(x-x_1)(x-x_2)\\\sf x^2+3x+2=1\cdot(x-(-1))\cdot(x-(-2))=(x+1)\cdot(x+2)\\\sf\dfrac{x-3}{x^2+3x+2}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$

$\boxed{\sf A=\dfrac{x-3}{x+2}\Bigg|_{x=-1}=\dfrac{-1-3}{-1+2}=-4}\\\boxed{\sf B=\dfrac{x-3}{x+1}\Bigg|_{x=-2}=\dfrac{-2-3}{-2+1}=\dfrac{-5}{-1}=5}\\\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx=-4\int\dfrac{dx}{x+1}+5\int\dfrac{dx}{x+2}\\\displaystyle\sf\int\dfrac{dx}{x+1}\\\underline{\rm fac_{\!\!,}a}\\\rm t=x+1\implies dt=dx\\\displaystyle\sf\int\dfrac{dx}{x+1}=\int\dfrac{dt}{t}=\ell n|t|+k\\\displaystyle\sf\int\dfrac{dx}{x+1}=\ell n|x+1|+k$

$\displaystyle\sf\int\dfrac{dx}{x+2}\\\underline{\rm fac_{\!\!,}a}\\\sf u=x+2\implies du=dx\\\displaystyle\sf\int\dfrac{dx}{x+2}=\int\dfrac{du}{u}=\ell n|u|+k\\\displaystyle\sf\int\dfrac{dx}{x+2}=\ell n|x+2|+k$

$\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx=-4\int\dfrac{dx}{x+1}+5\int\dfrac{dx}{x+2}\\\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx=-4\ell n|x+1|+5\ell n|x+2|+k\\\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx=\ell n|x+1|^{-4}+\ell n|x+2|^5+k\\\large\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\dfrac{x-3}{x^2+3x+2}~dx=\ell n\left|\dfrac{(x+2)^5}{(x+1)^4}\right|+k }\checkmark}}}$

$\huge\boxed{\boxed{\boxed{\boxed{\sf Espero~ter~ajudado~^-~_{\cup}~^-}}}}$