Question

Integrala de la 1 la 3 din 1/x^2-16 dx

Answer 1

$\it \ I = \int_0 ^1 \dfrac{1}{x^2-16} dx=?$

[tex]\it \dfrac{1}{x^2-16} = \dfrac{1}{x^2-4^2} = \dfrac{1}{(x-4)(x+4)} =\dfrac{A}{x-4} + \dfrac{B}{x+4} = \\\;\\ \\\;\\ = \dfrac{Ax+4A+Bx-4B}{(x-4)(x+4)} = \dfrac{(A+B)x +4(A-B)}{(x-4)(x+4)} \Longrightarrow [/tex]

[tex]\it \Longrightarrow \begin{cases}\it A+B=0 \Rightarrow A= -B\ \ \ (*) \\\;\\ \it 4(A-B) =1 \Rightarrow A-B = \dfrac{1}{4} \stackrel{(*)}{\Longrightarrow} -B-B=\dfrac{1}{4}\Rightarrow B = -\dfrac{1}{8}\end{cases} \\\;\\ \\\;\\ \Longrightarrow A = \dfrac{1}{8}[/tex]

[tex]\it I = \int^1_0 \dfrac{\dfrac{1}{8}}{x-4}dx - \int^1_0 \dfrac{\dfrac{1}{8}}{x+4}dx = \dfrac{1}{8}\left[(ln|x-4|)\big|^1_0 -(ln|x+4|)\big|^1_0\right] = \\\;\\ \\\;\\ \dfrac{1}{8} (ln3-ln4-ln5+ln4) = \dfrac{1}{8} ln\dfrac{3}{5}[/tex]