Antiderivatives And Indefinite Integration

ANTIDERIVATIVES AND INTEGRATION
Before you begin the exercises, be sure you realize that one of the most important steps in integration is rewriting the integrand in a form that fits the basic integration rules. To illustrate this point, here are some examples. Original Integral Rewrite Integrate Simplify

∫
∫ (t
2

2 dx x + 1 dt

2∫ x

−1 2

dx

)

2

∫ (t

4

+ 2t 2 + 1dt
-2

)

 x 12   2  1 +C  2 5 t3  t + 2  + t + C 3 5  

4x

1 2

+C

1 5 2 3 t + t +t +C 5 3
1 2 3 x − +C 2 x

x3 + 3 ∫ x 2 dx

∫ (x + 3x ) dx
 x 4 3 − 4 x 13  dx  ∫  

∫

3

x (x − 4 ) dx

 x −1  x2 + 3  −1  +C  2   7  x 43  x 3  − 4  4 +C 7   3  3

3 43 x (x − 7 ) + C 7

Complete the table as the examples of the top of thispage as a model.

Original Integral

Rewrite

Integrate

Simplify

∫

3

x dx 1
2

∫x ∫ x (x ∫x
2

dx dx

1 x 1
3

+ 3 dx dx
2

)

∫ 2x

∫ (3x )

1

dx

Find the indefinite integral. Larson, page 283. 1. 2. 3.

∫ (x + 3) dx ∫ (x
3

+ 5 dx
3 2

)

 ∫x 

+ 2x + 1 dx  

1 2 x + 3x + C 2 1 = x 4 + 5x + C 4 2 52 = x + x2 + x +C 5
=

4. 5. 6.7.
8. 9. 10. 11. 12. 13. 14. 15.

∫x

1
3

dx

=− =

1 +C 2x 2

x2 + x + 1 ∫ x dx

∫ (x + 1)(3x - 2) dx
2 ∫ y y dy

2 12 2 x 3 x + 5x + 15 + C 15 1 = x 3 + x 2 − 2x + C 2
2 7 = y 2 +C 7

(

)

∫ dx ∫ (2 sin x + 3 cos x ) dx ∫ (1 − csc t cot t ) dt ∫ (2 sin x − 5e )dx ∫ (sec θ − sinθ )dθ ∫ (tan y + 1) dy
x

= x +C = −2 cos x + 3 sin x + C = t + csc t + C

= −2 cos x −5e x + C = tanθ + cos θ + C = tan y + C

2

2

∫ (2x − 4 )dx
x

∫  x − x  dx  



5

4x =x − +C ln 4 1 = x 2 − 5 ln x + C 2
2

Integration by substitution. First rule.
Complete the table by identify u and du for the integral. Larson, página 305

∫ f (g ( x )) g ' ( x ) dx ∫ (5x
2

u = g(x)

du = g’(x) dx

+ 1 (10 x ) dx

)

∫x
∫

2

x 3 + 1 dx
x dx

x+1
2

∫ sec 2x tan 2x dx ∫ sec ∫ sin
nca
2

x tan 2 x dx

cos x dx 3 x

2 of 17

Find the indefinite integral. 16. 17. 18. 19. 20. 21. 22.

∫ 2(1 + 2x )

4

dx

(1 + 2x )5 =
=

∫ (− 2x )
∫ x (x
3 4

9 − x 2 dx
+ 3 dx − 1 dx

)

2

(x = (
(t =

2 9 − x2 3
4

(

5

+C

)

3

2

+C

∫ x (x
2

3

)

4

+3 +C 12 5 x3 −1 = +C 15

)

3)

∫t ∫ 5x

t + 2 dt
2
3

1 − x 2 dx
2 3

∫ (1 − x )
x
2

x

dx

+2 2 +C 3 4 15 = − 1− x 2 3 + C 8 1 = +C 2 4 1− x 2
2

)

3

(

)

(

)

23. 24. 25. 26. 27.
28. 29.

∫ (1 + x )
∫
x 1− x
2

3 2

dx
dx

=−

1 +C 3 1+ x 3

(

)

= − 1− x 2 + C 1  1 = − 1 +  + C 4 t
4

 1  1  ∫ 1 + t   t 2  dt     1 ∫ 2x dx x 2 + 3x + 7dx ∫ x 2 2 ∫ t  t − t  dt  

3

= 2x + C
3 1 2 5 = x 2 + 2 x 2 + 14 x 2 + C 5

∫ (9 − y )

y dy

1 = t4 −t2 +C 4 3 2 5 = 6y 2 − y 2 + C 5
Larson, page 332.

30. 31. 32.

∫ π sin πx dx ∫ sin 2x dx ∫θ
1
2

= − cos πx + C 1 = − cos 2x + C 2 1 = − sin  + C θ  = e 5x + C

1 cos  dθ θ  33. ∫ 5e 5 x dx

nca

3 of 17

34. 35. 36. 37.

2 −x ∫ x e dx
3

∫sin 2x cos 2x dx ∫ tan
4

x sec 2 xdx

csc 2 x ∫ cot 3 x dx 38. ∫ cot 2 x dx
x x ∫ e e + 1 dx

1 3 = − e −x + C 3 1 2 1 sin 2 x + C or − cos 2 2x + C 4 4 1 5 = tan x + C 5 1 2 1 tan x + C or sec 2 x + C 2 2 = − cot x − x + C
= =−
3 1 x e +1 +C 3

39. 40.

(

)

2

(

)

∫e

x

1 - e x dx

5 −ex 41. ∫ 2 x dx e
42. 43. 44. 45.

∫e ∫e ∫3

sin πx
−x x

cos πx dx
23 2 1− e x 2 + C 3 5 = − e −2 x + e − x + C 2 1 sin πx = e +C

(

)

sec e dx

( ) dx
−x

π

= − tan e − x + C
2  x2  3  + C  ln 3  2 1 =− 5 −x + C 2 ln 5

( )

2

=

∫ x5

−x2

dx

( )

POOL OF COMBINED EXERCISES 46. 47. 48. 49. 50. 51. 52. 53.

∫ sin10xdx ∫ sec (1 − 4 x )dx
2

∫ cos ∫ cos ∫

4

x sin xdx x dx

2

1 − 4 x dx

∫ (5x − 1)
5...