Integración por fracciones parciales ejemplo 10


Esto es lo que usted escribira:

1 / ((x + 1) (x^3 + 1))

∫ 1/((x + 1) (x^3 + 1)) d x
<br />          Factorizando el denominador en términos lineales y <br />          términos quadraticos irreducibles . <br />         
= ∫ 1/((x + 1)^2 (x^2 - x + 1)) d x
<br />          Separando las fracciones en fracciones parciales . <br />         
= ∫ ((1 - x)/(3 (x^2 - x + 1)) + 1/(3 (x + 1)) + 1/(3 (x + 1)^2)) d x
<br />          Integrando la suma término-por-término y factorizando las constantes . <br />         
= 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x + 1/3 ∫ (1 - x)/(x^2 - x + 1) d x
<br />          Completando el cuadrado in (1 - x)/(x^2 - x + 1) . <br />         
= 1/3 ∫ (1 - x)/((x - 1/2)^2 + 3/4) d x + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          Para el integrante (1 - x)/((x - 1/2)^2 + 3/4), <br />          sustituya s = x - 1/2, <br />          d s = 1 d x . <br />         
= 1/3 ∫ (1/2 - s)/(s^2 + 3/4) d s + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          Para el integrante (1/2 - s)/(s^2 + 3/4), <br />          sustituya t = (2 s)/3^(1/2), <br />          d t = 2/3^(1/2) d s . <br />         
= 2/(3 3^(1/2)) ∫ (1/2 - (3^(1/2) t)/2)/(t^2 + 1) d t + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          Integrando la suma (1/2 - (3^(1/2) t)/2)/(t^2 + 1) téermino-por-término . <br />         
= 1/(3 3^(1/2)) ∫ 1/(t^2 + 1) d t - 1/3 ∫ t/(t^2 + 1) d t + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          La integral de 1/(t^2 + 1) es tan^(-1)(t) . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/3 ∫ t/(t^2 + 1) d t + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          Para el integrante t/(t^2 + 1), <br />          sustituya w = t^2 + 1, <br />          d w = 2 t d t . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 ∫ 1/w d w + 1/3 ∫ 1/(x + 1)^2 d x + 1/3 ∫ 1/(x + 1) d x
<br />          Para el integrante 1/(x + 1)^2, <br />          sustituya y = x + 1, <br />          d y = 1 d x . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 ∫ 1/w d w + 1/3 ∫ 1/(x + 1) d x + 1/3 ∫ 1/y^2 d y
<br />          La integral de 1/y^2 es -1/y . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 ∫ 1/w d w + 1/3 ∫ 1/(x + 1) d x - 1/(3 y)
<br />          Para el integrante 1/(x + 1), <br />          sustituya y = x + 1, <br />          d y = 1 d x . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 ∫ 1/w d w + 1/3 ∫ 1/y d y - 1/(3 y)
<br />          La integral de 1/y es log(y) . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 ∫ 1/w d w + log(y)/3 - 1/(3 y)
<br />          La integral de 1/w es log(w) . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - log(w)/6 + log(y)/3 - 1/(3 y) + ÷r
<br />          Resustituyendo y = x + 1 . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - log(w)/6 + 1/3 log(x + 1) - 1/(3 (x + 1)) + ÷r
<br />          Resustituyendo w = t^2 + 1 . <br />         
= tan^(-1)(t)/(3 3^(1/2)) - 1/6 log(t^2 + 1) + 1/3 log(x + 1) - 1/(3 (x + 1)) + ÷r
<br />          Resustituyendo t = (2 s)/3^(1/2) . <br />         
= tan^(-1)((2 s)/3^(1/2))/(3 3^(1/2)) - 1/6 log((4 s^2)/3 + 1) + 1/3 log(x + 1) - 1/(3 (x + 1)) + ÷r
<br />          Resustituyendo s = x - 1/2 . <br />         
= tan^(-1)((2 x - 1)/3^(1/2))/(3 3^(1/2)) + 1/3 log(x + 1) - 1/6 log(1/3 (2 x - 1)^2 + 1) - 1/(3 (x + 1)) + ÷r
<br />          Factor por otra expresión para ver el resultado . <br />         
= (2 3^(1/2) x tan^(-1)((2 x - 1)/3^(1/2)) + 2 3^(1/2) tan^(-1)((2 x - 1)/3^(1/2)) + 6 x log(x + 1) + 6 log(x + 1) - 3 x log(1/3 (2 x - 1)^2 + 1) - 3 log(1/3 (2 x - 1)^2 + 1) - 6)/(18 (x + 1)) + ÷r

Converted by Mathematica  (March 14, 2003)