Problemas de Integración por Fracciones Parciales MA-II ccesa007
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Problemas de Integración por Fracciones Parciales MA-II ccesa007
- 2. Ejemplo 1 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥
- 3. 2𝑥2+3 𝑥2+1 2 = 𝐴𝑥+𝐵 𝑥2+1 + 𝐶𝑥+𝐷 𝑥2+1 2 2𝑥2+3 𝑥2+1 2 = 𝐴𝑥3+𝐴𝑥+𝐵𝑥2+𝐵+𝐶𝑥+𝐷 𝑥2+1 2 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 2𝑥2 + 3 = 𝐴𝑥3 + 𝐴𝑥 + 𝐵𝑥2 + 𝐵 + 𝐶𝑥 + 𝐷
- 4. 2𝑥2 + 3 = 𝐴𝑥3 + 𝐵𝑥2 + 𝐴 + 𝐶 𝑥 + 𝐵 + 𝐷 𝐴 = 0 𝐵 = 2 𝐴 + 𝐶 = 0 𝐵 + 𝐷 = 3 → 0 + 𝐶 = 0 → 𝐶 = 0 → 2 + 𝐷 = 3 → 𝐷 = 1
- 5. 2𝑥2+3 𝑥2+1 2 = 𝐴𝑥+𝐵 𝑥2+1 + 𝐶𝑥+𝐷 𝑥2+1 2 2𝑥2+3 𝑥2+1 2 = 2 𝑥2+1 + 1 𝑥2+1 2 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 2𝑑𝑥 𝑥2 + 1 + 𝑑𝑥 𝑥2 + 1 2 𝐴 = 0 𝐵 = 2 𝐶 = 0 𝐷 = 1
- 6. 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 2𝑑𝑥 𝑥2 + 1 + 𝑑𝑥 𝑥2 + 1 2 Integral Original: 2𝑑𝑥 𝑥2 + 1 = 2 𝑑𝑥 𝑥2 + 1 tan 𝑧 = 𝑥 1 𝑑𝑥 = sec2 𝑧 𝑑𝑧 tan 𝑧 = 𝑥 sec 𝑧 = 𝑥2 + 1 1 sec 𝑧 = 𝑥2 + 1 sec2 𝑧 = 𝑥2 + 1 z 1 x
- 7. 𝑑𝑥 = sec2 𝑧 𝑑𝑧 sec2 𝑧 = 𝑥2 + 1 tan 𝑧 = 𝑥 2𝑑𝑥 𝑥2 + 1 = 2 𝑑𝑥 𝑥2 + 1 2𝑑𝑥 𝑥2 + 1 = 2 tan−1 𝑥 + 𝐶 = 2 sec2 𝑧 𝑑𝑧 sec2 𝑧 = 2 𝑑𝑧 = 2𝑧 + C
- 8. 𝑑𝑥 = sec2 𝑧 𝑑𝑧 sec2 𝑧 = 𝑥2 + 1 𝑑𝑥 𝑥2 + 1 2 = sec2 𝑧 𝑑𝑧 sec2 𝑧 2 = 1 sec2 𝑧 𝑑𝑧 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 2𝑑𝑥 𝑥2 + 1 + 𝑑𝑥 𝑥2 + 1 2 Integral Original:
- 9. sec 𝑥 = 1 cos 𝑥 Identidad Trigonométrica: → cos 𝑥 = 1 sec 𝑥 1 sec2 𝑧 𝑑𝑧 = cos2 𝑧 𝑑𝑧 = cos 𝑧 ∙ cos 𝑧 𝑑𝑧 𝑢 = cos 𝑧 𝑑𝑢 = − sin 𝑧 𝑑𝑧 𝑑𝑣 = cos 𝑧 𝑑𝑧 𝑑𝑣 = cos 𝑧 𝑑𝑧 𝑣 = sin 𝑧 = cos 𝑧 ∙ sin 𝑧 − sin 𝑧 − sin 𝑧 𝑑𝑧 = cos 𝑧 ∙ sin 𝑧 + sin2 𝑧 𝑑𝑧
- 10. Identidad Trigonométrica: sin2 𝑥 = 1 − cos2 𝑥cos2 𝑧 𝑑𝑧 = cos 𝑧 ∙ sin 𝑧 + (1 − cos2 z) 𝑑𝑧 cos2 𝑧 𝑑𝑧 = cos 𝑧 ∙ sin 𝑧 + 𝑑𝑧 − cos2z 𝑑𝑧 cos2 𝑧 𝑑𝑧 = cos 𝑧 ∙ sin 𝑧 + 𝑧 − cos2z 𝑑𝑧 2 cos2 𝑧 𝑑𝑧 = cos 𝑧 ∙ sin 𝑧 + 𝑧 cos2 𝑧 𝑑𝑧 = 1 2 (cos 𝑧 ∙ sin 𝑧) + 1 2 𝑧 + 𝐶
- 11. cos2 𝑧 𝑑𝑧 = 1 2 (cos 𝑧 ∙ sin 𝑧) + 1 2 𝑧 + 𝐶 = 1 2 𝑥 𝑥2 + 1 + 1 2 tan−1 𝑥 + 𝐶 z 1 x tan 𝑧 = 𝑥 = 1 2 1 𝑥2 + 1 ∙ 𝑥 𝑥2 + 1 + 1 2 tan−1 𝑥 + 𝐶 𝑑𝑥 𝑥2 + 1 2 = 1 2 𝑥 𝑥2 + 1 + 1 2 tan−1 𝑥 + 𝐶
- 12. 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 2𝑑𝑥 𝑥2 + 1 + 𝑑𝑥 𝑥2 + 1 2 𝑑𝑥 𝑥2 + 1 2 = 1 2 𝑥 𝑥2 + 1 + 1 2 tan−1 𝑥 + 𝐶 2𝑑𝑥 𝑥2 + 1 = 2 tan−1 𝑥 + 𝐶 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 2 tan−1 𝑥 + 1 2 𝑥 𝑥2 + 1 + 1 2 tan−1 𝑥 + 𝐶 2𝑥2 + 3 𝑥2 + 1 2 𝑑𝑥 = 5 2 tan−1 𝑥 + 1 2 𝑥 𝑥2 + 1 + 𝐶
- 13. Ejemplo 2 • Resolver sin 𝑥 𝑑𝑥 cos 𝑥 1+cos2 𝑥 , sea cos 𝑥 = 𝑢
- 14. 𝑢 = cos 𝑥 𝑥 = cos−1 𝑢 𝑑𝑥 = − 1 1 − 𝑢2 𝑑𝑢 Identidad Trigonométrica: sin2 𝑥 + cos2 𝑥 = 1 sin2 𝑥 + 𝑢2 = 1 sin 𝑥 = 1 − 𝑢2 sin 𝑥 𝑑𝑥 cos 𝑥 1 + cos2 𝑥 = 1 − 𝑢2 𝑢 1 + 𝑢2 ∙ − 1 1 − 𝑢2 𝑑𝑢 Fórmula: 𝑑 𝑑𝑥 tan−1 𝑣 = − 𝑑𝑣 𝑑𝑥 1 − 𝑣2
- 15. 1 − 𝑢2 𝑢 1 + 𝑢2 ∙ − 1 1 − 𝑢2 𝑑𝑢 = − 1 𝑢 1 + 𝑢2 1 𝑢 1 + 𝑢2 = 𝐴 𝑢 + 𝐵𝑢 + 𝐶 1 + 𝑢2 1 𝑢 1 + 𝑢2 = 𝐴 + 𝐴𝑢2 + 𝐵𝑢2 + 𝐶𝑢 𝑢 1 + 𝑢2 1 = 𝐴 + 𝐵 𝑢2 + 𝐶𝑢 + 𝐴 𝐶 = 0 𝐴 = 1 𝐴 + 𝐵 = 0 → 𝐵 = −1 − 1 𝑢 1 + 𝑢2 = − 1 𝑢 𝑑𝑢 + −1𝑢 1 + 𝑢2 𝑑𝑢
- 16. − 𝑑𝑢 𝑢 − 𝑢 𝑑𝑢 1 + 𝑢2 = 𝑢 𝑑𝑢 1 + 𝑢2 − 𝑑𝑢 𝑢 = 1 2 ln 1 + 𝑢2 − ln 𝑢 + 𝐶 = ln 1 + 𝑢2 1 2 − ln 𝑢 + 𝐶 = 1 2 2𝑢 𝑑𝑢 1 + 𝑢2 − 𝑑𝑢 𝑢
- 17. sin 𝑥 𝑑𝑥 cos 𝑥 1 + cos2 𝑥 = ln 1 + cos2 𝑥 cos 𝑥 + 𝐶 Integral Original: sin 𝑥 𝑑𝑥 cos 𝑥 1 + cos2 𝑥 cos 𝑥 = 𝑢 − 1𝑑𝑢 𝑢 1 + 𝑢2 = ln 1 + 𝑢2 ln 𝑢 + 𝐶
- 18. Ejemplo 3 2 + tan2 𝜃 sec2 𝜃 𝑑𝜃 1 + tan3 𝜃
- 19. 2 + tan2 𝜃 sec2 𝜃 𝑑𝜃 1 + tan3 𝜃 𝑢 = tan 𝜃 Identidad Trigonométrica: 1 + tan2 𝜃 = sec2 𝜃 1 + u2 = sec2 𝜃 𝜃 = tan−1 𝑢 𝑑𝜃 = 1 1 + u2 𝑑𝑢 = 2 + u2 (1 + u2 ) 1 + u3 ∙ 1 1 + u2 𝑑𝑢 = 2 + u2 𝑑𝑢 1 + u3 = 2 + u2 𝑑𝑢 (1 + 𝑢) 1 − 𝑢 + 𝑢2
- 20. 2 + u2 (1 + 𝑢) 1 − 𝑢 + 𝑢2 = 𝐴 1 + 𝑢 + 𝐵𝑢 + 𝐶 1 − 𝑢 + 𝑢2 = 𝐴 + 𝐵 𝑢2 + −𝐴 + 𝐵 + 𝐶 𝑢 + 𝐴 + 𝐶 (1 + 𝑢)(1 − 𝑢 + 𝑢2) 2 + u2 = 𝐴 + 𝐵 𝑢2 + −𝐴 + 𝐵 + 𝐶 𝑢 + 𝐴 + 𝐶 𝐴 + 𝐵 = 1 −𝐴 + 𝐵 + 𝐶 = 0 𝐴 + 𝐶 = 2 𝐴 = 1 𝐵 = 0 𝐶 = 1
- 21. 𝐴 = 1 𝐵 = 0 𝐶 = 1 2 + u2 𝑑𝑢 (1 + 𝑢) 1 − 𝑢 + 𝑢2 = 𝐴 1 + 𝑢 𝑑𝑢 + 𝐵𝑢 + 𝐶 1 − 𝑢 + 𝑢2 𝑑𝑢 = 1 1 + 𝑢 𝑑𝑢 + 𝑑𝑢 1 − 𝑢 + 𝑢2 = ln(1 + 𝑢) + 𝑑𝑢 𝑢2 − 𝑢 + 1 4+ 3 4 = ln(1 + 𝑢) + 𝑑𝑢 𝑢 − 1 2 2 + 3 2 2
- 22. z 3 2 𝑢 − 1 2 tan 𝑧 = 2𝑢 − 1 3 𝑢 = 3 tan 𝑧 + 1 2 𝑑𝑢 = 3 2 sec2 𝑑𝑧 sec 𝑧 = 2 ∙ 𝑢 − 1 2 2 + 3 2 2 3 3 4 𝑠𝑒𝑐2 𝑧 = 𝑢 − 1 2 2 + 3 2 2
- 23. 𝑑𝑢 = 3 2 sec2 𝑑𝑧 3 4 sec2 𝑧 = 𝑢 − 1 2 2 + 3 2 2 𝑧 = tan−1 2𝑢 − 1 3 𝑑𝑢 𝑢 − 1 2 2 + 3 2 2 = 3 2 sec2 𝑑𝑧 3 4sec2 𝑧 = 2 3 3 𝑑𝑧 = 2 3 3 𝑧 + 𝐶 2 + u2 𝑑𝑢 (1 + 𝑢) 1 − 𝑢 + 𝑢2 = ln(1 + 𝑢) + 𝑑𝑢 𝑢 − 1 2 2 + 3 2 2 2 + u2 𝑑𝑢 (1 + 𝑢) 1 − 𝑢 + 𝑢2 = ln(1 + 𝑢) + 2 3 3 tan−1 2𝑢 − 1 3 + 𝐶
- 24. Integral Original: 2 + tan2 𝜃 sec2 𝜃 𝑑𝜃 1 + tan3 𝜃 𝑢 = tan 𝜃 2 + u2 𝑑𝑢 (1 + 𝑢) 1 − 𝑢 + 𝑢2 = ln(1 + 𝑢) + 2 3 3 tan−1 2𝑢 − 1 3 + 𝐶 2 + tan2 𝜃 sec2 𝜃 𝑑𝜃 1 + tan3 𝜃 = ln 1 + tan 𝜃 + 2 3 3 tan−1 2 tan 𝜃 − 1 3
- 25. Ejemplo 4 𝑡 + 3 𝑡2 + 4𝑡 + 5 2 𝑑𝑡
- 26. 𝑡 + 3 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 = 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 = 𝐴𝑡 + 𝐵 𝑡2 + 4𝑡 + 5 + 𝐶𝑡 + 𝐷 𝑡2 + 4𝑡 + 5 2 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 = (𝐴𝑡 + 𝐵) 𝑡2 + 4𝑡 + 5 + 𝐶𝑡 + 𝐷 𝑡2 + 4𝑡 + 5 2 𝑡2 + 6𝑡 + 9 = 𝐴𝑡3 + 4𝐴𝑡2 + 5𝐴𝑡 + 𝐵𝑡2 + 4𝐵𝑡 + 5𝐵 + 𝐶𝑡 + 𝐷
- 27. 𝑡2 + 6𝑡 + 9 = 𝐴𝑡3 + 4𝐴 + 𝐵 𝑡2 + 5𝐴 + 4𝐵 + 𝐶 𝑡 + (5𝐵 + 𝐷) 𝐴 = 0 4𝐴 + 𝐵 = 1 → 𝐵 = 1 5𝐴 + 4𝐵 + 𝐶 = 6→ 𝐷 = 45𝐵 + 𝐷 = 9 → 𝐶 = 2 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 = 𝐴𝑡 + 𝐵 𝑡2 + 4𝑡 + 5 + 𝐶𝑡 + 𝐷 𝑡2 + 4𝑡 + 5 2 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 = 1 𝑡2 + 4𝑡 + 5 + 2𝑡 + 4 𝑡2 + 4𝑡 + 5 2
- 28. 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 = 𝑑𝑡 𝑡2 + 4𝑡 + 5 + (2𝑡 + 4) 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 = 𝑑𝑡 𝑡 + 2 2 + 1 + (2𝑡 + 4) 𝑡2 + 4𝑡 + 5 −2 𝑑𝑡 = 𝑑𝑡 𝑡 + 2 2 + 1 + 𝑡2 + 4𝑡 + 5 −1 −1 = 𝑑𝑡 𝑡 + 2 2 + 1 − 1 𝑡2 + 4𝑡 + 5
- 29. 𝑑𝑡 𝑡 + 2 2 + 1 z 1 𝑡 + 2 tan 𝑧 = 𝑡 + 2 1 𝑑𝑡 = sec2 𝑧 𝑑𝑧 tan 𝑧 = 𝑡 + 2 sec 𝑧 = 𝑡 + 2 2 + 1 1 sec 𝑧 = (𝑡 + 2)2+1 sec2 𝑧 = (𝑡 + 2)2+1 𝑡 = tan 𝑧 − 2
- 30. 𝑑𝑡 = sec2 𝑧 𝑑𝑧 sec2 𝑧 = (𝑡 + 2)2+1 tan 𝑧 = 𝑡 + 2 𝑑𝑡 𝑡 + 2 2 + 1 = sec2 𝑧 𝑑𝑧 sec2 𝑧 = 𝑑𝑧 = 𝑧 + 𝐶 𝑑𝑡 𝑡 + 2 2 + 1 = tan−1 𝑡 + 2 + 𝐶
- 31. 𝑑𝑡 𝑡 + 2 2 + 1 = tan−1 𝑡 + 2 + 𝐶 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 = 𝑑𝑡 𝑡 + 2 2 + 1 − 1 𝑡2 + 4𝑡 + 5 𝑡2 + 6𝑡 + 9 𝑡2 + 4𝑡 + 5 2 𝑑𝑡 = tan−1 𝑡 + 2 − 1 𝑡2 + 4𝑡 + 5 + 𝐶