Question

Resolva a integral:

$\large\displaystyle\begin{aligned}\int \frac{5x-3}{x^2-2x-3}\ dx\end{aligned}$

Resposta:

2 Ln| x-3 | + 3 Ln| x+1 | + k

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Produto~de~Stevin}\\\sf (x+a)\cdot(x+b)=x^2+(a+b)x+a\cdot b\\\sf x^2-2x-3=x^2+(1-3)+1\cdot(-3)\\\sf x^2-2x-3=(x+1)\cdot(x-3)\\\sf\dfrac{5x-3}{x^2-2x-3}=\dfrac{5x-3}{(x+1)\cdot(x-3)}=\dfrac{A}{x+1}+\dfrac{B}{x-3}\\\sf A=\dfrac{5x-3}{x-3}\bigg |_{x=-1}=\dfrac{5\cdot(-1)-3}{-1-3}=\dfrac{-8}{-4}=2\\\\\sf B=\dfrac{5x-3}{x+1}\bigg |_{x=3}=\dfrac{5\cdot3-3}{3+1}=\dfrac{12}{4}=3\\\\\sf\dfrac{5x-3}{x^2-2x-3}=\dfrac{2}{x+1}+\dfrac{3}{x-3}\end{array}}$

$\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{5x-3}{x^2-2x-3}dx=\int\dfrac{2}{x+1}dx+\int\dfrac{3}{x-3}dx\\\displaystyle\sf\int\dfrac{2}{x+1}dx=2\int\dfrac{dx}{x+1}\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf k=x+1\implies dk=dx\\\displaystyle\sf2\int\dfrac{dx}{x+1}=2\int\dfrac{dk}{k}=2\ell n|k|+c_1\\\displaystyle\sf2\int\dfrac{dx}{x+1}=2\ell n|x+1|+c_1\\\displaystyle\sf\int\dfrac{3}{x-3}dx=3\int\dfrac{dx}{x-3}\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf z=x-3\implies dz=dx\\\displaystyle\sf3\int\dfrac{dx}{x-3}=3\int\dfrac{dz}{z}=3\ell n|z|+c_2\\\displaystyle\sf3\int\dfrac{dx}{x-3}=3\ell n|x-3|+c_2\end{array}}$

$\Large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{5x-3}{x^2-2x-3}dx=2\ell n|x+1|+3\ell n|x-3|+c_1+c_2\\\displaystyle\sf\int\dfrac{5x-3}{x^2-2x-3}dx=\ell n\bigg|(x+1)^2\cdot(x-3)^3\bigg|+k,onde~k=c_1+c_2\end{array}}$

Answer 2

Resposta: $2ln(|x+1|)+3ln(|x-3|)+k$

Observe que, fatorando o denominador por agrupamento:

$\displaystyle\int\dfrac{5x-3}{x^2-2x-3}\,dx$

$\displaystyle\int\dfrac{5x-3}{x^2+x-3x-3}\,dx$

$\displaystyle\int\dfrac{5x-3}{x(x+1)-3(x+1)}\,dx$

$\displaystyle\int\dfrac{5x-3}{(x+1)(x-3)}dx$

Pode-se reescrever essa fração numa soma de frações parciais:

$\displaystyle\int\dfrac{5x-3}{(x+1)(x-3)}dx=\displaystyle\int\dfrac{k_1}{x+1}+\dfrac{k_2}{x-3}dx$

$\dfrac{5x-3}{(x+1)(x-3)}=\dfrac{k_1}{x+1}+\dfrac{k_2}{x-3}$ ⇒ isole somente ''5x – 3''.

$5x-3=(x+1)(x-3)\bigg(\dfrac{k_1}{x+1}+\dfrac{k_2}{x-3}\bigg)$

$5x-3=(x+1)(x-3)\bigg(\dfrac{k_1}{x+1}\bigg)+(x+1)(x-3)\bigg(\dfrac{k_2}{x-3}\bigg)$

$5x-3=(x-3)k_1+(x+1)k_2$

$5x-3=k_1x-3k_1+k_2x+k_2$

$5x-3=(k_1+k_2)x-3k_1+k_2$

$\begin{cases}5=k_1+k_2\\-\,3=-\,3k_1+k_2\end{cases}$

Note que para k₁ = 2 e k₂ = 3 as igualdades são válidas:

$\begin{cases}5=2+3\\-\,3=-\,3(2)+3\end{cases}\Leftrightarrow~~\begin{cases}5=5\\-\,3=-\,3\end{cases}$

Daí:

$\displaystyle\int\dfrac{k_1}{x+1}+\dfrac{k_2}{x-3}dx~\Leftrightarrow~\displaystyle\int\dfrac{2}{x+1}+\dfrac{3}{x-3}dx$

A integral da soma é igual a soma das integrais de cada parcela:

$\displaystyle\int\dfrac{2}{x+1}dx+\displaystyle\int\dfrac{3}{x-3}dx$

Reescreva as integrais tomando a x + 1 = u e x – 3 = v:

$\displaystyle\int\dfrac{2}{u}du+\displaystyle\int\dfrac{3}{v}dv$

Basta lembrar que 1/x é a derivada de ln(|x|), então ln(|x|) é a integral de 1/x; com base nisso:

$2ln(|u|)+c_1+3ln(|v|)+c_2$ ⇒ (some ao final de cada integração as constantes arbitrárias).

Retome a u = x + 1 e v = x – 3:

$2ln(|x+1|)+c_1+3ln(|x-3|)+c_2$ ⇒ c₁ + c₂ = k

$2ln(|x+1|)+3ln(|x-3|)+k$

PORTANTO:

$\displaystyle\int\dfrac{5x-3}{x^2-2x-3}\,dx=2ln(|x+1|)+3ln(|x-3|)+k$

Bons estudos e um forte abraço. — lordCzarnian9635.