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

$\large\boxed{\begin{array}{l}\underline{\rm Integrac_{\!\!,}\tilde ao~de~func_{\!\!,}\tilde oes~racionais}\\\underline{\rm por~frac_{\!\!,}\tilde oes~parciais}\\\sf\dfrac{2x+1}{x^2-1}=\dfrac{2x+1}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}\\\sf A=\dfrac{2x+1}{x+1}\bigg |_{x=1}=\dfrac{2\cdot1+1}{1+1}=\dfrac{3}{2}\\\\\sf B=\dfrac{2x+1}{x-1}\bigg|_{x=-1}=\dfrac{2\cdot(-1)+1}{-1-1}=\dfrac{1}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\displaystyle\sf\int\dfrac{2x+1}{x^2-1}~dx=\dfrac{3}{2}\ell n|x-1|+\dfrac{1}{2}\ell n|x+1|+k\\\displaystyle\sf\int\dfrac{2x+1}{x^2-1}~dx=\ell n\bigg|(x-1)^{\frac{3}{2}}\cdot(x+1)^{\frac{1}{2}}\bigg|+k\end{array}}$

Answer 2

Explicação passo-a-passo:

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

$\Large \boxed{\underline{\bf \blue{Bons\: Estudos!}}}\\ \\\Large \boxed{\underline{\bf 28/05/2021}}$