Question

⚠️Ajuda?? Calcule o valor da integral indefinida usando o método para decompor em frações parciais.

teorema: ∫ $\frac{P(x)}{(x-\alpha )(x-\beta )} =$ A ln$|x-\alpha | + B ln |x-\beta |+k$

Answer 1

$\int{\frac{x^3+x+1}{x^2-1}dx}$

Simplificando o quociente e utilizando as frações parciais:

$\int{x+\frac{2x+1}{x^2-1}dx}\\\\\int{x+\frac{2x+1}{(x+1)(x-1)}dx}\\\\\int{x+\frac{A}{x+1}+\frac{B}{x-1}dx}\\\\\frac{x^2}{2}+A.\ln|x+1|+B.\ln|x-1|+k$

Calculando o valor dos coeficientes:

$2x+1=A(x-1)+B(x+1)\\\\2x+1=Ax-A+Bx+B\\\\2x+1=(A+B)x-A+B\\\\\left \{ {{A+B=2} \atop {B-A=1}} \right. \\\\B=\frac{3}{2}\\\\A=\frac{1}{2}$

Portanto

$\int{\frac{x^3+x+1}{x^2-1}dx}=\frac{x^2}{2}+\frac{1}{2}\ln|x+1|+\frac{3}{2}\ln|x-1|+k$

Ou ainda

$\frac{1}{2}(x^2+\ln|x+1|+3\ln|x-1|)+k\\\\\frac{1}{2}[x^2+\ln(|x+1|.|x-1|^3)]+k\\\\\frac{1}{2}[x^2+\ln(|(x+1)(x-1)^3|)]+k\\\\\frac{1}{2}[x^2+\ln(|(x^2-1)(x-1)^2|)]+k$