1 CÁLCULO DE PRIMITIVAS 2 1.1. INTEGRALES INMEDIATAS. EJEMPLOS Instrucciones de uso: tápese la solución antes de empezar a hacer la integral. Después de resuelta, compruébese que es correcta (sólo después). 1. R 2. R 3. R 4. R 5. R 6. R 7. R 8. 9. 1 dx (2x+1)2 = −1 2(2x+1) 2x+1 dx (x2 +x+1)2 1 dx x2 +2x+1 = = x dx 3x2 +1 = −1 2(x+1)2 3x2 +1 3 +C +C √ 6(x+1)7/6 x+1 √ +C 3 x+1 dx = 7 √ √ x+1 dx = 2 x + 1 + C x+1 R √ x2 1 + x3 dx = R √ x 1 − x2 dx = 10. +C −1 x+1 √ = +C −1 x2 +x+1 1 dx x3 +3x2 +3x+1 √ +C R √ 2(1+x3 )3/2 9 −(1−x2 )3/2 3 2(1+x)3/2 3 1 + xdx = 27. R 28. R 29. R √ cos( x) √ dx 2 x = sen( x) + C 30. R cos(ln x) dx x = sen(ln x) + C 31. R cos(tg x) dx cos2 x = sen(tg x) + C 32. R cos(arc tg x) dx 1+x2 = sen(arc tg x) + C 33. R 1 dx 1+(3x+27)2 = 34. R 35. R e dx 1+e2x 36. R sec2 x 1+tg2 x =x+C 37. R 2x dx 1+4x = √ 1 dx x(1+x) +C +C +C cos(−x + 1)dx = − sen(−x + 1) + C 3 cos(2x + 6)dx = 3 x dx 1+x8 x = 3 2 arc tg(3x+27) 3 4 arc tg(x ) 4 arc tg(2x ) ln 2 R 1 dx 3x+5 = 1 3 38. R 12. R ln |3x + 5| + C 3 dx ax+b = 3 a ln |ax + b| + C 39. R 1 dx x(1+(ln x)2 ) 13. R x dx x3 +2 = 1 3 41. R 2x2 dx 6x3 +1 1 dx x2 +2x+2 14. R ln |x3 42. 1 dx 9+x2 = 1 3 15. R R e dx 1+ex 43. 1 dx 3+x2 = sen x−cos x dx sen x+cos x 44. 1 dx 4x2 +4x+2 17. R R 1 √ 3 16. R R 1 dx x ln x 45. 18. R R 1 dx (1+x2 ) arc tg x 46. 3 sec2 (2x + 6)dx = 19. R R e2x+1 dx 47. R x sec2 (x2 )dx = 20. R 48. R 21. R 49. R 22. R etg x sec2 xdx = etg x + C 50. R 23. R 5x 9x dx = 45x ln 45 +C 51. R 24. R earc tg x dx 1+x2 = earc tg x + C 52. R 25. R sen(arc tg x) dx 1+x2 e2x+1 dx 53. R 2x+1 dx x2 +x−6 26. R 54. R x = 1 9 + 2| + C ln |6x3 + 1| + C = ln(1 + ex ) + C = − ln | sen x+cos x|+C = ln(ln x) + C = ln(arc tg x) + C 1 2x+1 e 2 = 2 +C 2 e−x xdx = − 12 e−x + C ex 3 +1 x2 dx = = 3 1 x +1 e 3 1 2x+1 e 2 +C +C 2x cos(x2 + 2)dx = sen(x2 + 2) + C +C +C = arc tg(ex ) + C 11. 2 sen(2x + 6) + C √ +C √ = 2 arc tg( x) + C = arc tg(ln x) + C = arc tg(x + 1) + C arc tg( x3 ) + C arc tg( √x ) + C 3 = 1 2 arc tg(2x + 1) + C sec2 (−x + 1)dx = − tg(−x + 1) + C 1 2 3 2 tg(2x + 6) + C tg(x2 ) + C sen(2x + 5)dx = − 12 cos(2x + 5) + C 2x sen(x2 + 2)dx = − cos(x2 + 2) + C sen(ln x) dx = − 21 cos(ln x) + C 2x √ √ sen( x) √ dx = −2 cos( x) + C x = − cos(arc tg x) + C = ln |x2 + x − 6| + C x−1 dx x2 −2x−6 = 1 2 ln |x2 − 2x − 6| + C