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PRACTICO Nº 2

ANÁLISIS MATEMÁTICO II Calcular las siguientes integrales

1.∫

𝑥 2 +1 2𝑥 3 +3𝑥

𝑑𝑥 = ∫ 2

𝑥 2 +1 2𝑥 3 +3𝑥 2

𝐴

𝑥 2 +1 𝑥 2 (2𝑥+3) 𝐵

𝑑𝑥

2𝑥 3 + 3𝑥 2 = 𝑥 2 (2𝑥 + 3)

𝐶

= 𝑥 + 𝑥 2 + 2𝑥+3

𝑥 2 +1 2𝑥 3 +3𝑥 2

=

𝐴𝑥(2𝑥+3)+𝐵(2𝑋+3)+𝐶𝑥 2 𝑥 2 (2𝑥+3)

𝑥 2 + 1 = 𝐴𝑥(2𝑥 + 3) + 𝐵(2𝑋 + 3) + 𝐶𝑥 2 𝑥=0

1

1 = 3𝐵 →

𝑥=1

𝐵=3

SUMANDO 1 Y 2 15𝐴 + 3𝐶 = 1 3𝐴 − 3𝐶 = −5 --------------------------18𝐴 = −4 𝐴 = − 2⁄9 REEEMPLAZANDO “A” EN 1 2 15 (− ) + 3𝐶 = 1 9 10 3+10 3𝐶 = 1 − 3 = 3

2 = 5𝐴 + 5𝐵 + 𝐶 5

2 = 5𝐴 + 3 + 𝐶 5

5𝐴 + 𝐶 = 2 − 3 15𝐴 + 3𝐶 = 1 … … … . . (1) 𝑥 = −1 2 = −𝐴 + 𝐵 + 𝐶 1 𝐴−𝐶 = −2 3

5

𝐴 − 𝐶 = −3 3𝐴 − 3𝐶 = −5 … … … . . (2)

𝐶=

2

𝑑𝑥 𝑥

+ ∫ 2+ 3 𝑥

2

𝑑𝑥 𝑥

+ 3 ∫ 𝑥 −2 𝑑𝑥 + 9∗2 ∫ 2𝑥+3

=− ∫ 9 = −9∫

1

𝑑𝑥

13 𝑑𝑥 ∫ 2𝑥+3 9

1

2

13

1

2𝑑𝑥

13

= − 9 ln|𝑥| + 3 (−𝑥 −1 ) + 18 ln|2𝑥 + 3| + 𝑐 13

2

1

= 18 ln|2𝑥 + 3| − 9 ln|𝑥| − 3𝑥 + 𝑐 3 3 2.∫ 𝑥2+3𝑥 𝑑𝑥 = ∫ 𝑑𝑥 𝑥(𝑥+3) 3 𝑥(𝑥+3) 3 𝑥(𝑥+3)

𝐴

𝑥 2 + 3𝑥 = 𝑥(𝑥 + 3)

𝐵

= 𝑥 + 𝑥+3 =

𝐴(𝑥+3)+𝐵𝑥 𝑥(𝑥+3)

3 = 𝐴(𝑥 + 3) + 𝐵𝑥 𝑥=0 𝑥 = −3

3 = 3𝐴 3 = −3𝐵

→ →

𝐴=1 𝐵 = −1

13 9

=∫

𝑑𝑥 𝑑𝑥 −∫ 𝑥 𝑥+3

= ln|𝑥| − ln|𝑥 + 3| + 𝑐 𝑥 = ln +𝑐 𝑥+3 3𝑥+1 3.∫ 𝑥23𝑥+1 𝑑𝑥 = ∫ (𝑥+1)(𝑥+3) 𝑑𝑥 +4𝑥+3 3𝑥+1 𝑥 2 +4𝑥+3

= 𝑥+1 + 𝑥+3

𝐴

3𝑥+1 𝑥 2 +4𝑥+3

=

𝑥 2 + 4𝑥 + 3 = (𝑥 + 1)(𝑥 + 3)

𝐵

𝐴(𝑥+3)+𝐵(𝑥+1) (𝑥+1)(𝑥+3)

3𝑥 + 1 = 𝐴(𝑥 + 3) + 𝐵(𝑥 + 1) 𝑥 = −1 𝑥 = −3 = −∫

−2 = 2𝐴 −8 = −2𝐵

→ →

𝐴 = −1 𝐵=4

𝑑𝑥 𝑑𝑥 + 4∫ 𝑥+1 𝑥+3

= −ln|𝑥 + 1| + 4 ln|𝑥 + 3| + 𝑐 = ln(𝑥 + 3)4 − ln|𝑥 + 1| + 𝑐 = ln

(𝑥 + 3)4 +𝑐 𝑥+1

4.∫ 𝑥35𝑥(𝑥2+3 𝑑𝑥 = +2) 3

5𝑥 3 +3 𝑥 3 (𝑥 2 +2) 5𝑥 3 +3 𝑥 3 (𝑥 2 +2)

𝑥 3 (𝑥 2 + 2) 𝐴 𝑥

= + =

𝐵 𝑥2

+

𝐶 𝑥3

𝐷𝑥+𝐸 +2)

+ (𝑥 2

𝐴𝑥 2 (𝑥 2 +2)+𝐵𝑥(𝑥 2 +2)+𝐶(𝑥 2 +2)+(𝐷𝑥+𝐸)𝑥 3 𝑥 3 (𝑥 2 +2)

5𝑥 3 + 3 = 𝐴𝑥 2 (𝑥 2 + 2) + 𝐵𝑥(𝑥 2 + 2) + 𝐶(𝑥 2 + 2) + (𝐷𝑥 + 𝐸)𝑥 3 𝑥=0 𝑥=1

33 = 2𝐶 → 𝐶 = 3⁄2 8 = 3𝐴 + 3𝐵 + 3𝐶 + 𝐷 + 𝐸 3 8 = 3𝐴 + 3𝐵 + 3 2 + 𝐸 9 2

3𝐴 + 3𝐵 + 𝐷 + 𝐸 = 8 − =

7 2

6𝐴 + 6𝐵 + 2𝐷 + 2𝐸 = 7 … … … . . (1) 𝑥 = −1 −2 = 3𝐴 − 3𝐵 + 3𝐶 + 𝐷 − 𝐸 3 3𝐴 − 3𝐵 + 𝐷 − 𝐸 = −2 − 3 2 9

3𝐴 − 3𝐵 + 𝐷 − 𝐸 = −2 − 2 3𝐴 − 3𝐵 + 𝐷 − 𝐸 = 3𝐴 − 3𝐵 + 𝐷 − 𝐸 =

−4−9 2 13 −2

6𝐴 − 6𝐵 + 2𝐷 − 2𝐸 = 13 ……….(2) 𝑥=2 43 = 24𝐴 + 12𝐵 + 6𝐶 + 16𝐷 + 8𝐸 3 24𝐴 + 12𝐵 + 16𝐷 + 8𝐸 = 43 − 6 ∗ 2 24𝐴 + 12𝐵 + 16𝐷 + 8𝐸 = 43 − 9

24𝐴 + 12𝐵 + 16𝐷 + 8𝐸 = 34 … … . (3) 𝑥 = −2 −37 = 24𝐴 − 12𝐵 + 6𝐶 + 16𝐷 − 8𝐸 24𝐴 − 12𝐵 + 16𝐷 − 8𝐸 = −37 − 9 24𝐴 − 12𝐵 + 16𝐷 − 8𝐸 = −46 … … . . (4) SUMANDO 1 Y 2 6𝐴 + 6𝐵 + 2𝐷 + 2𝐸 = 7 6𝐴 − 6𝐵 + 2𝐷 − 2𝐸 = 13 ---------------------------------------12𝐴 + 4𝐷 = 20……….(5)

SUMANDO 3 Y 4

43 12 REEEMPLAZANDO “A” EN 7 𝐷 = (20 − 12𝐴)/4 𝐷 = 5 − 3𝐴 3 ∗ 43 𝐷 =5− 12 𝐷 = 72 − 129 𝐷 = −57

24𝐴 + 12𝐵 + 16𝐷 + 8𝐸 = 34 24𝐴 − 12𝐵 + 16𝐷 − 8𝐸 = −46 ----------------------------------------------48𝐴 + 32𝐷 = −12 … … . . (6) 𝑫𝑬 (𝟓) 12𝐴 4𝐷 = 20 𝐷 = (20 − 12𝐴)/4………..(7) (7)𝑅𝐸𝐸𝑀𝑃𝐿𝐴𝑍𝐴𝑀𝑂𝑆𝐸𝑁 (6) 48𝐴 32𝐷 = −12 48𝐴 + 32(5 − 3𝐴) = −12 48𝐴 + 160 − 96𝐴 = −12 −48𝐴 = −172 𝐴

𝐵

𝐶

𝐴=

LOS VALORES DE B Y E SON CEROS

𝐷𝑥+𝐸

= 𝑥 + 𝑥 2 + 𝑥 3 + (𝑥2 +2) =

43 𝑑𝑥 3 𝑑𝑥 𝑥𝑑𝑥 ∫ + ∫ 3 + (−57) ∫ 2 12 𝑥 2 𝑥 𝑥 +2

=

43 3 57 2𝑥 𝑑𝑥 ln|𝑥| + ∫ 𝑥 −3 𝑑𝑥 − ∫ 2 12 2 2 𝑥 +2

43 3 𝑥 −2 57 ln|𝑥| + − ln|𝑥 2 + 2| + 𝑐 12 2 −2 2 43 3 57 = ln|𝑥| − 2 − ln|𝑥 2 + 2| + 𝑐 12 4𝑥 2 =

5.∫ 𝑥22𝑥+7 𝑑𝑥 = +2𝑥−4

𝑥 2 + 2𝑥 − 4 → −2+√22 −4∗1∗(−4) 2∗1 −2+√20 2

; ;

−1 + √5 ; 2𝑥+7 𝑥(𝑥+3) 2𝑥+7 𝑥(𝑥+3)

= =

−𝑏±√𝑏2 −4𝑎𝑐 2𝑎 −2−√22 −4∗1∗(−4) 2∗1 −2−√20 2

−1 − √5

𝐴 𝐵 + (𝑥+1+√5) (𝑥+1−√5) 𝐴(𝑥+1+√5)+𝐵(𝑥+1−√5) (𝑥+1−√5)(𝑥+1+√5)

2𝑥 + 7 = 𝐴𝑥 + 𝐴 + 𝐴√5 + 𝐵𝑥 + 𝐵 − 𝐵√5 2𝑥 + 7 = 𝑥(𝐴 + 𝐵) + 𝐴 + 𝐴√5 + 𝐵 − 𝐵√5 𝑥=0 𝑥 = −3

2= 𝐴+𝐵 7 = 𝐴 + 𝐴√5 + 𝐵 − 𝐵√5 7 = 𝐴√5 − 𝐵√5 + 𝐴 + 𝐵 5 = 𝐴√5 − 𝐵√5

2√5 = 𝐴√5 + 𝐵 √5 5 = 𝐴√5 − 𝐵√5 −−−−−−−−−− 2√5 + 5 = 2𝐴√5

𝐴 =1+

𝐴=

5 2√5

2 + √5 2

REEMPLAZANDO “A” 2=𝐴+𝐵 𝐵 =2−𝐴 2 + √5 𝐵 =2− 2

=

∫

2𝑥 + 7 𝐴 𝐵 𝑑𝑥 = ∫ 𝑑𝑥 + ∫ 𝑑𝑥 𝑥(𝑥 + 3) (𝑥 + 1 − √5) (𝑥 + 1 + √5)

=

2 + √5 𝑑𝑥 2 − √5 𝑑𝑥 ∫ + ∫ 2 2 (𝑥 + 1 − √5) (𝑥 + 1 + √5)

𝐵=

4 − 2 − √5 2

𝐵=

2 − √5 2

2 + √5 2 − √5 ln|𝑥 + 1 − √5| + ln|𝑥 + 1 + √5| + 𝑐 2 2

5𝑥−7 6. ∫ (𝑥2+2)(𝑥−3) 𝑑𝑥 =

5𝑥−7 (𝑥 2 +2)(𝑥−3)

=

5𝑥−7 (𝑥 2 +2)(𝑥−3)

𝐴𝑥+𝐵 𝐶 + (𝑥 2 +2) (𝑥−3)

=

(𝐴𝑥+𝐵)(𝑥−3)+𝐶(𝑥 2 +2) (𝑥 2 +2)(𝑥−3)

5𝑥 − 7 = (𝐴𝑥 + 𝐵)(𝑥 − 3) + 𝐶(𝑥 2 + 2) 𝑥=3

8= 11𝐶 →

𝐶= 16

𝑥=0

8 11

𝑥=1

−7 = −3𝐵 + 11 16

77

3𝐵 = 11 + 11 93

3𝐵 = 11 93

𝐵 = 33

=∫ =∫

→

31

𝐵 = 11

𝐴𝑥 + 𝐵 𝐶 𝑑𝑥 + ∫ 𝑑𝑥 2 (𝑥 + 2) (𝑥 − 3)

𝐴𝑥 𝐵 𝐶 𝑑𝑥 + ∫ 2 𝑑𝑥 ∫ 𝑑𝑥 + 2) (𝑥 + 2) (𝑥 − 3)

(𝑥 2

=−

8 2∗𝑥 31 1 8 1 ∫ 2 𝑑𝑥 + ∫ 2 𝑑𝑥 + ∫ 𝑑𝑥 11 ∗ 2 (𝑥 + 2) 11 (𝑥 + 2) 11 (𝑥 − 3)

=−

8 31 1 𝑥 8 ln|𝑥 2 + 2| + ∗ arctan + ln|𝑥 − 3| + 𝑐 22 11 √2 √2 11

−2 = −2𝐴 − 2𝐵 + 3𝐶 2𝐴 = 3𝐶 − 2𝐵 + 2 8 31 −2∗ +2 11 11 24 62 22 2𝐴 = − + 11 11 11 16 2𝐴 = − 11 8 𝐴=− 11

2𝐴 = 3 ∗

=− =

8 31√2 𝑥√2 8 ln|𝑥 2 + 2| + arctan + ln|𝑥 − 3| + 𝑐 22 22 2 11

8 8 31√2 𝑥√2 ln|𝑥 − 3| − ln|𝑥 2 + 2| + arctan ++𝑐 11 22 22 2

𝑑𝑥 7. ∫ 9𝑥𝑑𝑥 4 +𝑥 2 = ∫ 𝑥 2 (9𝑥 2 +1)

9𝑥 4

9𝑥 4 + 𝑥 2 = 𝑥 2 (9𝑥 2 + 1)

1 𝐴 𝐵 𝐶𝑥 + 𝐷 = + 2+ 2 +𝑥 𝑥 𝑥 (9𝑥 2 + 1)

1 𝐴𝑥(9𝑥 2 + 1) + 𝐵(9𝑥 2 + 1) + (𝐶𝑥 + 𝐷)𝑥 2 = 9𝑥 4 + 𝑥 2 𝑥 2 (9𝑥 2 + 1) 1 = 𝐴𝑥(9𝑥 2 + 1) + 𝐵(9𝑥 2 + 1) + (𝐶𝑥 + 𝐷)𝑥 2 1 = 9𝐴𝑥 3 + 𝐴𝑥 + 9𝐵𝑥 2 + 𝐵 + 𝐶𝑥 3 + 𝐷𝑥 2 1 = (9𝐴 + 𝐶)𝑥 3 + (9𝐵 + 𝐷)𝑥 2 + 𝐴𝑥 + 𝐵 0 = 9𝐴 + 𝐶

→

𝐶=0

0 = 9𝐵 + 𝐷

→

𝐷 = −9

0=𝐴

→

𝐴=0

1=B

→

𝐵=1

𝐴 𝐵 𝐶𝑥 + 𝐷 = ∫ 𝑑𝑥 + ∫ 2 𝑑𝑥 + ∫ 𝑑𝑥 𝑥 𝑥 (9𝑥 2 + 1) 0 1 0𝑥 + (−9) = ∫ 𝑑𝑥 + ∫ 2 𝑑𝑥 + ∫ 𝑑𝑥 𝑥 𝑥 (9𝑥 2 + 1) = ∫ 𝑥 −2 𝑑𝑥 − 9 ∫

𝑑𝑥 9𝑥 2 + 1

𝑥 −1 1 3𝑥 − 9 ∗ arctan + 𝑐 −1 1 1 1 = − − 9 arctan 3𝑥 + 𝑐 𝑥 =