Question

Calcule o valor da integral indefinida usando o método para 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

Explicação passo-a-passo:

$\Large \int \frac{2x + 1}{ {x}^{2} - 1} dx \\ \\ \frac{2x + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} \\ \\ \frac{2x + 1 = (x + 1)A + (x - 1)B}{(x - 1)(x + 1) \: } \\ \\ 2x + 1 =Ax + A + Bx - B \\ \\ 2x + 1 = Ax + Bx + A- B \\ \\ 2x + 1 = (A+ B)x + (A- B) \\ \\ \left\{ \:{{A+ B = 2} \atop{A - B = 1}} \right. \\ \\ 2A = 3 \\ \\ \Large \boxed{ A = \frac{2}{3} } \\ \\ A+ B = 2 \\ \\ \frac{2}{3} + B = 2 \\ \\ B = - \frac{3}{2} + 2 \\ \\ B = \frac{ - 1 \times 3 + 2 \times 2}{2} \\ \\ B = \frac{ - 3 + 4}{2} \\ \\ \Large\boxed{B = \frac{1}{2} } \\ \\ \frac{A}{x - 1} + \frac{B}{x + 1} \\ \\ \frac{ \frac{3}{2} }{x - 1} + \frac{ \frac{1}{2} }{x + 1} \\ \\ \frac{3}{2} \times \frac{1}{x - 1} + \frac{1}{2} \times \frac{1}{x + 1} \\ \\ \Large \int\frac{3}{2(x - 1)} + \frac{1}{2(x + 1)} dx \\ \\\Large \int\frac{3}{2(x - 1)} dx + \int\frac{1}{2(x + 1)} dx \\ \\\Large\boxed{ \green{\frac{3}{2} ln |x - 1| + \frac{1}{2} ln |x + 1| + k, \: k \in \mathbb{R}}}$

$\Large \boxed{\underline{\bf \blue{Bons\: Estudos!}}}\\ \\\Large \boxed{\underline{\bf 27/05/2021}}$