Primitivas

Primitivas con soluci´n o 2x4 + 3x3 − x2 dx = x2 + 3x − ln |x| + k x3 3 5 9x2 + 5 dx = x2 + ln |3x| + k 3x 2 3 √ 1 (3x2 + 2) 2x3 + 4x + 1dx = (2x3 + 4x + 1)3/2 + k 3 √ 4 √ 2 dx =(x2 + 1)3/4 + k 3 x2 + 1 x

1. 2. 3. 4. 5.

3 6√ dx = − 4 − 5x + k 5 4 − 5x
2

6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

1 xex 2 dx = ln |ex − 2| + k x2 − 2 2 e 3x dx= 1 x 3 +k ln(3)
2 /4

2x(7 − ex

dx = 7x2 − 4ex

2 /4

+k

xe+1 ex2 e + + e ln |x| + k (ex + xe + ex + )dx = ex + x e+1 2 2 6x2 − 11x + 5 dx = x3 − 3x + ln |3x − 1| +k 3x − 1 3 (2x − 1)(x + 3) dx = x2 + 15x + 72 ln |x − 5| + k x−5 3e2x 3 = ln |e2x + 1| + k 2x + 1 e 2 e7/x 1 dx = − e7/x + k 2 x 7 2x3 dx = −x2 + 4 ln |x2 − 4| + k 2−4 x 5 − 4x2dx = −x2 + 3x − 2 ln |2x + 3| + k 3 + 2x √ 2 √ ( x + 2)2 √ = ( x + 2)3 + k 9 3 x 3ex 3 dx = ln |6 + 5ex | + k x 6 + 5e 5

1

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29.30. 31. 32. 33. 34. 35.

√ √ 1+ x 4 √ dx = (1 + x)3/2 + k 3 x ln(x) (ln(x))2 dx = +k x 2 √ √ 2 (5 − t3/2 )1 4 t(5 − t t)0 4 dt = − +k 3 14 ln2 (x + 1) ln3 (t + 1) dx = +k x+1 38x3 − 6x2 − ex4 ex2 8 +k dx = x − 2 ln |x| − 3x3 3 6 2 dx = ln(ln(2x2 )) + k x ln(2x2 ) xex (ex + 3)−1/2 dx = (ex + 3)+ k x+3 dx = x − 3 ln |x + 6| + k x+6 (xe2x + 2x)dx = x 2x 12x e − e + x2 + k 2 4
2 2 2

x2 x3 + x2 − x − 3 dx = + x + ln |x2 − 3| + k x2 − 3 2 5x − 2 dx = 2 ln |x| + 3 ln |x − 1| + k x2 − x 3x3 − 3x + 4 3 1 x+1 +k dx = x2 + ln 2−4 4x 82 x−1 x3 17x − 12 dx = ln |x| + 2 ln |x − 4| − 3 ln |x + 3| + k − x2 − 12x

4−x 4 3 5 dx = + ln |x| + ln |x − 1| − ln |x + 1| + k 4 − x2 x x 2 2 1 1 dx = (ln |x − 1| − ln |x +1|) + k x2 − 1 2 1 dx = ln |x − 2| − ln |x − 3| + k (x − 2)(x − 3) 1 1 dx = (ln |x − 2| − ln |x + 3|) + k (x − 2)(x + 3) 5 1 dx = ln |x − 1| − ln |x|) + k x(x − 1)

2