Question

Calcule o valor da integral indefinida usando o método pra decompor em frações parciais.
teorema a ser usado: ∫$\frac{P(x)}{(x-\alpha)(x-\beta ) }$= A In $|x-\alpha |+BIn|x-\beta |+k$

Answer 1

$\displaystyle \int \frac{\text x^2+3\text x+1}{\text x^2-2\text x-3}\text {dx} \\\\\\ \int \frac{\text x^2-2\text x-3+5\text x+4}{\text x^2-2\text x-3}\text {dx} \\\\\\ \int (\frac{5\text x+4}{\text x^2-2\text x-3}+\frac{\text x^2-2\text x-3}{\text x^2-2\text x-3})\text{dx} \\\\\\ \int\frac{5\text x+4}{\text x^2-2\text x-3}\text{dx}+\int 1 \text{dx} \\\\\\ \int \frac{5\text x+4}{(\text x+1)(\text x-3)}\text{dx} + \text x\\\\\\ \underline{\text{Fa{\c c}amos} }:$

$\displaystyle \frac{5\text x+4}{(\text x+1)(\text x-3)}= \frac{\text A}{(\text x+1)}+\frac{\text B}{(\text x-3)} \\\\\\ \frac{5\text x+4}{(\text x+1)(\text x-3)} = \frac{\text {Ax}-3\text A+\text {Bx}+\text B}{(\text x+1)(\text x-3)} \\\\\\ \frac{5\text x +4}{(\text x+1)(\text x-3) } = \frac{(\text{A+B})\text x-3\text A+\text B}{(\text x+1)(\text x-3) } \\\\\\ \left \{ {{\text{A+B}=5} \atop {-3\text {A+ B}=4}} \right. \to 4\text A = 1 \to \boxed{\text A = \frac{1}{4} \ ; \ \text B = \frac{19}{4}}$

Daí :

$\displaystyle \int \frac{5\text x+4}{(\text x+1)(\text x-3)}\text {dx}+\text x = \int \frac{1}{4(\text x+1)}\text{dx}+\int \frac{19}{4(\text x-3)}\text{dx} +\text x \\\\\\ \frac{1}{4}\int \frac{\text {dx}}{\text x+1}+\frac{19}{4}\int \frac{\text{dx}}{\text x-3} \\\\\\\ \underline{\text{Portanto}} : \\\\ \int \frac{\text x^2+3\text x+1}{\text x^2-2\text x-3}\text{dx}= \boxed{\frac{1}{4}.\text{ln }|\text x+1| + \frac{19}{4}.\text{ln }|\text x-3| + \text x +\text C\ }\checkmark$  

Answer 2

Resultado da integral

$\Large \boxed{\boxed{ \sf\dfrac{19}{4} ln | x - 3 | + \dfrac{1}{4} ln | x + 1 | + k }}$

• Com K € aos reais

Temos a Seguinte Integral indefinida:

$\Large \boxed{\boxed{\displaystyle\int \sf \dfrac{x^{2}+3x+1}{x^{2} -2x-3} dx }}$

Com o Teorema:

$\displaystyle \int \sf \dfrac{P(x)}{(x-\alpha)(x-\beta ) }= A \: In |x-\alpha |+B \: In|x-\beta |+k$

Vamos lá, Primeiramente temos que fatorar o Denominador, Veja Abaixo

$\displaystyle\int \sf \dfrac{x^{2}+3x+1}{x^{2} -2x-3} dx = \displaystyle\int \sf \dfrac{x^{2}+3x+1}{(x - 3).(x + 1)} dx$

Queremos encontar o valor de A e B, vamos começar a decompor a integral por frações parciais.

$\Large\sf \dfrac{x^{2}+3x+1}{(x - 3).(x + 1)} = \dfrac{A}{(x - 3)} + \dfrac{B}{(x + 1)}$

• Primeiramente Multiplicamos o Denominador da fração A com os termos da integral e depois Multiplicamos com o Denominador da fração B

$\large \boxed{ \begin{array}{c} \\ \sf \dfrac{x^2+3x+1}{(x - 3).(x + 1)} = \dfrac{A}{(x - 3)} + \dfrac{B}{(x + 1)} \\ \\\sf \dfrac{(x - 3).x^2+3x+1}{(x - 3).(x + 1)} = \dfrac{A}{(x - 3).(x - 3)} + (x - 3).\dfrac{B}{(x + 1)} \\ \\\sf \dfrac{ \cancel{(x - 3)}.x^2+3x+1}{ \cancel{(x - 3)}.(x + 1)} = \dfrac{A}{ \cancel{(x - 3).(x - 3)}} + (x - 3). \dfrac{B}{(x + 1)} \\ \\\sf \dfrac{x^2+3x+1}{(x + 1)} = A + (x - 3).\dfrac{B}{(x + 1)} \\ \: \end{array}}$

Temos x - 3, Logo x = 3, vamos substituir as Incógnitas x por 3

$\large \boxed{\begin{array}{c} \\ \sf \dfrac{x^2+3x+1}{(x + 1)} = A + (x - 3).\dfrac{B}{(x + 1)} \\\\\sf \dfrac{3^2+3.3+1}{(3 + 1)} = A + (3 - 3).\dfrac{B}{(3 + 1)} \\\\ \sf \dfrac{3^2+10}{4} = A + 0.\dfrac{B}{(4)} \\\\ \sf \dfrac{19}{4} = A \\\: \end{array}}$

• Achando valor de B:

$\large \boxed{ \begin{array}{c} \\ \sf \dfrac{x^2+3x+1}{(x - 3).(x + 1)} = \dfrac{A}{(x - 3)} + \dfrac{B}{(x + 1)} \\ \\ \sf \dfrac{(x + 1).x^2+3x+1}{(x - 3).(x + 1)} = \dfrac{A}{(x - 3)} + \dfrac{B}{(x + 1).(x + 1)} \\ \: \end{array}}$

$\large \boxed{\begin{array}{c} \\ \sf \dfrac{ \cancel{(x + 1)}.x^2+3x+1}{(x - 3). \cancel{(x + 1)}} = \dfrac{A}{ x - 3} + \dfrac{B}{ \cancel{(x + 1) .(x + 1)}} \\ \\\sf \dfrac{x^2+3x+1}{(x - 3)} = (x + 1).\dfrac{A}{x - 3} + B \\ \: \end{array}}$

Temos x + 1, Logo x = -1, vamos substituir as Incógnitas x por -1

$\large \boxed{ \begin{array}{c} \\ \sf \dfrac{x^2+3x+1}{(x - 3)} = (x + 1).\dfrac{A}{x - 3} + B \\ \\ \sf \dfrac{ (- 1)^2+3( - 1)+1}{( - 1- 3)} = ( - 1 + 1).\dfrac{A}{ - 1- 3} + B \\ \\ \sf \dfrac{ 1 - 3+1}{ - 4} = 0.\dfrac{A}{ - 4} + B \\ \\ \sf \dfrac{ 1 - 3+1}{ - 4} = + B \\ \\ \sf \dfrac{ - 1}{ - 4} = B \\ \\ \sf \dfrac{1}{ 4} = B \\\: \end{array}}$

Achamos o valor de A e B, vamos lá na nossa Integral indefinida e refaze-la, aplicando algumas manipulações Álgebricas. Veja Abaixo

$\large \begin{array}{c} \\ \displaystyle\int \sf \Bigg( \frac{A}{x - 3} + \dfrac{B}{x + 1} \Bigg)dx \\ \\\displaystyle\int \sf \Bigg( \frac{19}{4(x - 3)} + \dfrac{1}{4(x + 1)} \Bigg) dx \\ \\ \displaystyle\int \Bigg( \sf \dfrac{19}{4(x - 3)}\Bigg)dx + \displaystyle\int \Bigg(\sf \dfrac{1}{4(x + 1)}\Bigg)dx \\ \\ \dfrac{19}{4} \displaystyle\int \Bigg( \sf \dfrac{dx}{(x - 3)}\Bigg) + \dfrac{1}{4} \displaystyle\int \Bigg( \sf \dfrac{dx}{(x + 1)}\Bigg) \\\: \end{array}$

Aplicando Teorema - Resposta:

$\Large \boxed{\boxed{ \sf\dfrac{19}{4} ln | x - 3 | + \dfrac{1}{4} ln | x + 1 | + k }}$

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