Integrales Zegarra

Mett ®

Ana Mar´ Albornoz R. ıa

M´todos de integraci´n e o

1

Integrales inmediates y sustituciones sencillas
1. tg 3 x sec2 x dx tg x sec x dx = 2. x2
4

Sea u = tg x ⇒ du = sec2 x dx u4 tg 4 x u du = +C = +C 4 4
3

3

2

(x3

+

1)7

dx dx =

Sea u = x3 + 1 ⇒ du = 3x2 dx 1 3 1 u7/4 = −4 −4 +C = +C 3/4 3 + 1)3/4 3u 3 (x

x2
4

3.

(x3 + 1)7 √ √ sen x 1 √ dxSea u = x ⇒ du = √ dx x 2 x √ √ sen x √ dx = 2 sen u du = −2 cos u + C = −2 cos x + C x 4x2 2x − 3 1 dx = − 12x + 1 4 4x2 8x − 12 dx − 12x + 1

4.

Sea u = 4x2 − 12x + 1 ⇒ du = 8x − 12 dx 2x − 3 1 1 dx = du = ln u + C = ln |4x2 − 12x + 1| + C 4x2 − 12x + 1 4 u 5. Arcsen x √ dx 1 − x2 Arcsen x √ dx = 1 − x2 6. x sen x2 cos x2 dx x sen x2 cos x2 dx = 7. √ Sea u = Arcsen x ⇒ du = √ u du = 1 dx 1 −x2

u2 Arcsen2 x +C = +C 2 2

Sea u = sen x2 ⇒ du = 2x cos x2 dx 1 2 u du = u2 sen2 x2 +C = +C 4 4

1 1 dx Sea u = ln x ⇒ du = dx x x 1 − ln2 x 1 1 √ √ dx = du = Arcsen u + C = Arcsen ln x + C 2x x 1 − ln 1 − u2 x x √ √ 1 dx x2 − 1 Arcsec x 1 dx = x2 − 1 Arcsec x Sea u = Arcsec x ⇒ du =
2

8.

x
2

√

1 dx x2 − 1

u du =

u Arcsec x +C = +C 2 2 aex/2 + be−x/2 dx aex/2 − be−x/29.

aex + b dx = aex − b

aex + b e−x/2 dx = aex − b e−x/2

Mett ®

Ana Mar´ Albornoz R. ıa

M´todos de integraci´n e o

2

1 Sea u = aex/2 − be−x/2 ⇒ du = (aex/2 + be−x/2 ) dx 2 x ae + b 1 1 1 dx = du = ln u + C = ln |aex/2 − be−x/2 | + C x−b ae 2u 2 u 10. Las siguientes integrales resultan inmediatas para cierto valor particular del n´mero racional r. Determine en cada caso elvalor de r y calcule la integral correspondiente. a) xr ex dx x2 ex dx = b) xr ln x dx x−1 ln x dx = c) xr sen √ x dx
3 3

r = 2, u = x3 ⇒ du = 3x2 dx 1 3 1 u 1 3 e + C = ex + C 3 3 1 r = −1, u = ln x ⇒ du = dx x eu du = u2 ln2 x ln x dx = u du = +C = +C x 2 2 √ 1 1 r = − u = x ⇒ du = √ 2 x dx √ sen u du = −2cos u + C = −2cos x + C

√ sen x √ dx = 2 x

Integraci´n por partes o
11. 2x dx 1+ x2 2x Sea u = Arcsen ∧ dv = dx ⇒ 1 + x2 1 2(1 + x2 ) − 2x ∗ 2x v = x ∧ du = dx = 2 (1 + x2 )2 2x 1 − 1+x2 Arcsen 1
1+2x2 +x4 − 4x2 (1+x2 )2

1
(1+x2 )2 − 4x2 (1+x2 )2

2 + 2x2 − 4x2 dx = (1 + x2 )2 1

2 − 2x2 1 + x2 2 − 2x2 dx = √ dx = (1 + x2 )2 1 − 2x2 + x4 (1 + x2 )2

2(1 − x2 ) dx = (1 − x2 )2 (1 + x2 )

2 2 dx ⇒ du = dx 2) (1 + x (1 + x2 ) Por partes queda: 2x 2x Arcsen dx = xArcsen − 2 1+x 1 + x2 2x x Arcsen − ln (1 + x2 ) + C 1 + x2 12. x Arctg (3x + 4) dx

2x dx = (1 + x2 )

Sea u = Arctg (3x + 4) ∧ dv = x dx ⇒ du =

x2 1 dx ∧ v = 1 + (3x + 4)2 2

Mett ®

Ana Mar´ Albornoz R. ıa Por partes queda: x Arctg (3x + 4) dx = x2 1 Arctg (3x + 4) − 2 2 = = = x2 1 Arctg (3x + 4) − 2 2 x2 1 Arctg (3x + 4) − 2 2 x2 Arctg (3x + 4) − 2

M´todos de integraci´n e o

3x2 1 dx = 2 1 + (3x + 4)2

x2 dx Dividiendo los polinomios de la integral 9x2 + 24x + 17
−24x−17 1 9 + 2 dx 9 9x + 24x + 17

1 1 − 9 2

−24x−17 9

9x2 + 24x + 17 9x2

dx

x2 x 1 Arctg (3x + 4) − + 2 18 18

24x + 17 dx + 24x + 17

x2 x 18 Arctg (3x + 4) − + = 2 18 24 x2 x 18 = Arctg (3x + 4) − + 2 18 24 x2 x 18 = Arctg (3x + 4) − + 2 18 24 x2 x 18 = Arctg (3x + 4) − + 2 18 24 ==

18x + 51 4 dx 2 + 24x + 17 9x 18x + 51 + 45 − 45 4 4 4 dx 2 + 24x + 17 9x 18x + 24 − 45 4 dx 2 + 24x + 17 9x 18x + 24 18 dx − 2 + 24x + 17 9x 24
45 4

9x2 + 24x + 17 9x2

dx

x2 x 18 18 45 Arctg (3x + 4) − + ln |9x2 + 24x + 17| − 2 18 24 24 4 x2 x 18 135 Arctg (3x + 4) − + ln |9x2 + 24x + 17| − 2 18 24 16

1 dx + 24x + 17

1 dx (3x + 4)2 + 1 1 dt +1

Sea t = 3x + 4 ⇒ dt = 3 dxx2 x 18 135 1 = Arctg (3x + 4) − + ln |9x2 + 24x + 17| − 2 18 24 16 3 x2 Arctg (3x + 4) − 2 x2 Arctg (3x + 4) − = 2 √ 13. x Arctg x2 − 1 dx = Sea u = Arctg x Arctg = √ √ x 18 + ln |9x2 + 24x + 17| − 18 24 x 18 + ln |9x2 + 24x + 17| − 18 24

(t)2

45 Arctg t + C 16 45 Arctg (3x + 4) + C 16

x2 − 1 ∧ dv = x dx ⇒ du = √ x2 Arctg x2 − 1 − 2 x √ dx 2 x2 − 1

1 x2 2x √ dx ∧ v = 1 + x2 − 1 2 x2...