Question

Ayuda con esta integralk

Answer 1

$\displaystyle\\I=\int\dfrac{dx}{(x^2-2x+5)^2}\\ \\ \\\text{Metodo de Hermite}\\ \\ \\I=\dfrac{Ax+B}{x^2-2x+5}+\int\dfrac{Cx+D}{x^2-2x+5}dx\\ \\\text{Derivamos ambos miembros}\\ \\ \\\dfrac{1}{(x^2-2x+5)^2}\equiv \dfrac{A(x^2-2x+5)-(Ax+B)(2x-2)}{(x^2-2x+5)^2}+\dfrac{Cx+D}{x^2-2x+5}$

$\dfrac{1}{(x^2-2x+5)^2}\equiv\dfrac{Cx^3 - x^2(A + 2C - D) - x(2B - 5C + 2D) + 5A + 2B + 5D}{(x^2-2x+5)^2}\\ \\\text{Comparamos numeradores y tenemos el siguiente sistema}\\\\\begin{cases}C=0\\A + 2C - D=0\\2B - 5C + 2D=0\\5A + 2B + 5D=1\end{cases}\\ \\\\A=\dfrac{1}{8}~~\wedge ~~B=\dfrac{1}{8}~~\wedge ~~C=0~~\wedge ~~D=\dfrac{1}{8}$

$\text{Hasta aqu\'i tenemos la integral}\\ \displaystyle\\I=\dfrac{x-1}{8(x^2-2x+5)}+\dfrac{1}{8}\int\dfrac{1}{x^2-2x+5}dx\\ \\\\\text{Sea }I_1= \int\dfrac{1}{x^2-2x+5}dx\\ \\ \\I_1=\int \dfrac{1}{(x-1)^2+2^2}dx=\dfrac{1}{2}\arctan \dfrac{x-1}{2}\\ \\ \\\boxed{I=\dfrac{x-1}{8(x^2-2x+5)}+\dfrac{1}{16}\arctan \dfrac{x-1}{2}+C}$