5. ∫𝑑𝑢

√𝑢2 +𝑎2 =

Sea u= atanϴ

du = a𝑠𝑒𝑐2ϴ

𝑢2+𝑎2 = 𝑎2 𝑡𝑎𝑛2ϴ+𝑎2

=𝑎2 (𝑡𝑎𝑛2ϴ+1)

=𝑎2 𝑠𝑒𝑐2ϴ

√𝑢2 + 𝑎2 = √𝑎2 + 𝑠𝑒𝑐2 𝛳

=asecϴ

∫𝑎𝑠𝑒𝑐2 𝛳𝑑𝛳

𝑎𝑠𝑒𝑐𝛳 = ∫ 𝑠𝑒𝑐𝛳𝑑𝛳 = In |𝑠𝑒𝑐𝛳 + 𝑡𝑎𝑛𝛳|

= In|√𝑢2 +𝑎2

𝑎| +

𝑢

𝑎

=In|𝑎√𝑢2 +𝑎2+ 𝑎𝑢

𝑎2 |

=In |√𝑢2 +𝑎2 +𝑢

𝑎|

=In|√𝑢2 + 𝑎2 + 𝑢| - In|𝑎|+C

=In|√𝑢2 + 𝑎2 + 𝑢|+C

4. ∫𝑥2

(𝑥2+4)2 𝑑𝑥

∫𝑥2

(𝑥2+4)2 = 𝐴𝑋 +𝐵

𝑋2 +4+

𝐶𝑋 +𝐷

(𝑋2 +4)2

𝑋2= (AX+B)(𝑋2+4)+(CX+D)

𝑋2= A𝑋3+4AX+B𝑋2+4B+CX+D

𝑋2=𝑋3(𝐴)+𝑋2(𝐵)+X(4A+C)+(4B+D)

A=0

B=1

4A+C=O; 4(0)+C=0, C=0

AB+D=0;4(1)+D=0, D=-4

∫𝑥2

(𝑥2+4) 2 = 0𝑋 +1

𝑋2 +4+

(0)𝑋−4

(𝑋2+4)2

=∫𝑑𝑥

𝑥2+4 - ∫

𝑑𝑥

(𝑥2+4)2

𝑥 = 2𝑡𝑎𝑛𝜃

𝑑𝑥 = 2𝑠𝑒𝑐2 𝜃𝑑𝜃

∫2𝑠𝑒𝑐2 𝜃𝑑𝜃

4𝑡𝑎𝑛2𝜃 + 4 − 4 ∫

2𝑠𝑒𝑐2𝜃𝑑𝜃

(4𝑡𝑎𝑛2𝜃 + 4)2

2 ∫𝑠𝑒𝑐2𝜃𝑑𝜃

4(𝑡𝑎𝑛2 𝜃 + 1) − 8 ∫

𝑠𝑒𝑐2𝜃𝑑𝜃

(4(𝑡𝑎𝑛2𝜃 + 1))2

1

2∫

𝑠𝑒𝑐2 𝜃𝑑𝜃

𝑠𝑒𝑐2 𝜃−

1

2∫

𝑠𝑒𝑐2𝜃𝑑𝜃

𝑠𝑒𝑐4 𝜃

1

2∫ 𝑑𝜃 −

1

2∫

𝑑𝜃

𝑠𝑒𝑐2 𝜃

1

2∫ 𝑑𝜃 −

1

2∫

𝑑𝜃1

𝑐𝑜𝑠2𝜃

1

2∫ 𝑑𝜃 −

1

2∫ 𝑐𝑜𝑠2 𝜃𝑑𝜃

1

2∫ 𝑑𝜃 −

1

2∫

1

2(1 + 𝑐𝑜𝑠2𝜃)𝑑𝜃

1

2∫ 𝑑𝜃 −

1

4∫ 𝑑𝜃 −

1

4∫ 𝑐𝑜𝑠2𝜃

1

4∫ 𝑑𝜃 −

1

4∫ 𝑐𝑜𝑠2𝜃

1

4(𝜃 −

1

2𝑠𝑒𝑛2𝜃) =

1

4(𝑡𝑎𝑛−1 (

𝑥

2)) −

1

8𝑠𝑒𝑛2 (𝑡𝑎𝑛−1

𝑥

2) + 𝐶

5.1

4)

∫𝑥2

(𝑥2 + 42)2𝑑𝑥

𝑥 = 2. tan 𝜃

𝑑𝑥 = 2. 𝑆𝑒𝑐2𝜃 𝑑𝜃

∫(2. tan 𝜃)2

((2. tan 𝜃)2 + 4)2 2. 𝑠𝑒𝑐2𝜃 𝑑𝜃

∫4. 𝑡𝑎𝑛2𝜃

(4. 𝑡𝑎𝑛2𝜃 + 4)2 2. 𝑆𝑒𝑐2𝜃 𝑑𝜃

8 ∫𝑡𝑎𝑛2𝜃. 𝑠𝑒𝑐2𝜃 𝑑𝜃

(4 (𝑡𝑎𝑛2𝜃 + 1))2

8 ∫𝑡𝑎𝑛2𝜃 . 𝑠𝑒𝑐2𝜃 . 𝑑𝜃

(4 . 𝑠𝑒𝑐2𝜃)2

1

2∫

𝑡𝑎𝑛2𝜃 . 𝑠𝑒𝑐2𝜃 𝑑𝜃

𝑠𝑒𝑐4𝜃

1

2∫

𝑡𝑎𝑛2𝜃 𝑑𝜃

𝑠𝑒𝑐2𝜃

1

2∫

𝑠𝑒𝑛2 𝜃

𝑐𝑜𝑠2 𝜃 1

𝑐𝑜𝑠2 𝜃

𝑑𝜃

1

2∫ 𝑠𝑒𝑛2𝜃 𝑑𝜃

1

2∫

1 − cos(2𝜃) 𝑑𝜃

2

1

4∫ 𝑑𝜃 −

1

4∫ cos(2𝜃) 𝑑𝜃

𝒖 = 2𝜃 ; 𝒅𝒖 = 2𝑑𝜃 ; 𝒅𝒖

𝟐= 𝑑𝜃

1

4. 𝜃 −

1

8∫ cos(𝑢). 𝑑𝑢

1

4. 𝜃 −

1

8 . sin(2𝜃) + 𝐶

1

4. 𝜃 −

1

4 . [sin(𝜃). cos( 𝜃) ] + 𝐶

𝒙 = 2. 𝑡𝑎𝑛 𝜃

𝑡𝑎𝑛−1 (𝑥

2) = 𝜽

1

4(tan−1 (

𝑥

2) − (

𝑥

√𝑥2 + 4 .

2

√𝑥2 + 4)) + 𝐶

𝟏

𝟒(𝒕𝒂𝒏−𝟏 (

𝒙

𝟐) − (

𝟐𝒙

𝒙𝟐 + 𝟒)) + 𝑪

2

𝑥

5)

∫𝑑𝑢

√𝑢2 + 𝑎2

Sol:

𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖ó𝑛: 𝑢 = 𝑎. tan 𝜃

𝑑𝑢 = 𝑎. 𝑠𝑒𝑐2𝜃 𝑑𝜃

∫𝑎. 𝑠𝑒𝑐2𝜃 𝑑𝜃

√(𝑎. 𝑡𝑎𝑛𝜃)2 + 𝑎2

𝑎 ∫𝑠𝑒𝑐2𝜃 𝑑𝜃

√𝑎2 (𝑡𝑎𝑛2𝜃 + 1)

𝑎

𝑎∫

𝑆𝑒𝑐2𝜃 𝑑𝜃

𝑠𝑒𝑐𝜃

∫ sec 𝜃 𝑑𝜃

ln|tan 𝜃 + sec 𝜃| + 𝐶

𝑢 = 𝑎. 𝑡𝑎𝑛𝜃

𝑢

𝑎 = tan 𝜃

ln |𝑢

𝑎 +

√𝑢2 + 𝑎2

𝑎| + 𝐶

ln |𝑎𝑢 +𝑎√𝑢2 +𝑎2

𝑎| + 𝐶

ln |𝑎(𝑢 + √𝑢2 + 𝑎2)

𝑎| + 𝐶

𝑙𝑛 |𝑢 + √𝑢2 + 𝑎2| + 𝐶

6.1

6)

∫(4𝑥 − 2)

𝑥3 − 𝑥2 − 2𝑥𝑑𝑥

∫(4𝑥 − 2)

𝑥(𝑥2 − 𝑥 − 2)𝑑𝑥

∫(4𝑥 − 2)

𝑥(𝑥 − 2)(𝑥 + 1)𝑑𝑥

𝒂

𝒖

4𝑥 − 2 = (𝑥)(𝑥 − 2)(𝑥 + 1) [𝐴

𝑥+

𝐵

𝑥 − 2+

𝐶

𝑥 + 1]

4𝑥 − 2 = 𝐴(𝑥 − 2)(𝑥 + 1) + 𝐵(𝑥)(𝑥 + 1) + 𝐶(𝑥)(𝑥 − 2)

4𝑥 − 2 = 𝐴(𝑥2 + 𝑥 − 2𝑥 − 2) + 𝐵(𝑥2 + 𝑥) + 𝐶( 𝑥2 − 2𝑥)

4𝑥 − 2 = (𝐴𝑥2 − 𝐴𝑋 − 2𝐴) + (𝐵𝑥2 + 𝐵𝑋) + (𝐶𝑥2 − 2𝐶𝑋)

0 = 𝐴 + 𝐵 + 𝐶

4 = −𝐴 + 𝐵 − 2𝐶

2 = −2𝐴

−2 = −2𝐴

−2

−2= 𝐴 ; 1 = 𝐴

𝟎 = 𝑨 + 𝑩 + 𝑪

𝟒 = −𝑨 + 𝑩 − 𝟐𝑪

𝟒 = 𝟐𝑩 − 𝑪

𝑪 = 𝟐𝑩 − 𝟒

𝟒 = −𝟏 + 𝑩 − 𝟐(𝟐𝑩 − 𝟒)

𝟒 = −𝟏 + 𝑩 − 𝟒𝑩 + 𝟖

𝟒 = 𝟕 − 𝟑𝑩

−𝟑

−𝟑= 𝑩 ; 𝟏 = 𝑩

𝑯𝒂𝒍𝒍𝒂𝒏𝒅𝒐 𝑪:

𝑪 = 𝟐(𝟏) − 𝟒; 𝑪 = 𝟐 − 𝟒; 𝑪 = −𝟐

∫𝒅𝒙

𝒙+ ∫

𝒅𝒙

(𝒙 − 𝟐) − 𝟐 ∫

𝒅𝒙

(𝒙 − 𝟏)

ln[x] + ln[x − 2] − 2 ln[x + 1] + C

ln[(x). (x − 2)] − ln[(x + 1)2] + C

ln [𝑥(𝑥 − 2)

(𝑥 + 1)2] + 𝐶

𝟔. 𝟏

𝟏𝟖) ∫𝟒𝒙 + 𝟏

𝟐𝒙 + 𝟏𝒅𝒙

𝑢 = 2𝑥 𝑑𝑢 = 2𝑥 𝑙𝑜𝑔(2)𝑑𝑥

=1

𝑙𝑜𝑔 (2)∫

𝑢2 + 1

𝑢(𝑢 + 1)𝑑𝑢

=1

𝑙𝑜𝑔 (2)∫ (−

2

𝑢 + 1+

1

𝑢+ 1)𝑑𝑢

= −1

𝑙𝑜𝑔 (2)∫ 1𝑑𝑢 +

1

𝑙𝑜𝑔 (2)∫ −

1

𝑢𝑑𝑢 +

1

𝑙𝑜𝑔 (2)∫

1

𝑢 + 1𝑑𝑢

𝑠 = 𝑢 + 1 𝑑𝑠 = 𝑑𝑢

= −2 𝑙𝑜𝑔(𝑠)

𝑙𝑜𝑔 (2)+

1

𝑙𝑜𝑔 (2)∫ 1𝑑𝑢 +

1

𝑙𝑜𝑔 (2)∫

1

𝑢𝑑𝑢

= −2 𝑙𝑜𝑔 (𝑠)

𝑙𝑜𝑔 (2)+

𝑙𝑜𝑔 (𝑢)

𝑙𝑜𝑔 (2)+

1

𝑙𝑜𝑔 (2)∫ 1𝑑𝑢

= −2 𝑙𝑜𝑔 (𝑠)

𝑙𝑜𝑔 (2)+

𝑢

𝑙𝑜𝑔 (2)+

𝑙𝑜𝑔 (𝑢)

𝑙𝑜𝑔 (2)+ 𝐶

=𝑢 + 𝑙𝑜𝑔(2𝑥) − 2 𝑙𝑜𝑔 (2𝑥 + 1)

𝑙𝑜𝑔 (2)+ 𝐶

=𝟐𝒙 + 𝒙 𝒍𝒐𝒈(𝟐) − 𝟐 𝒍𝒐𝒈 (𝟐𝒙 + 𝟏)

𝒍𝒐𝒈 (𝟐)+ 𝑪

𝟔. 𝟐

𝟔) ∫ (𝒅𝒙

𝟗𝒙𝟒 + 𝒙𝟐)

= ∫ (1

𝑥2) − (

9

9𝑥4 + 1)

= ∫ (1

𝑥2) 𝑑𝑥 − ∫ (

1

9𝑥2 + 1)𝑑𝑥

𝑢 = 3𝑥 𝑑𝑢 = 3𝑑𝑥

= ∫1

𝑥2𝑑𝑥 − 3 ∫

1

𝑢2 + 1𝑑𝑢

= ∫1

𝑥2𝑑𝑥 − 3 tan−1 𝑢

= ∫1

𝑥2𝑑𝑥 − 3 tan−1 𝑢

−1

𝑥− 3 tan−1 𝑢 + 𝐶

−𝟏

𝒙− 𝟑 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝑪

𝟏𝟖)

∫𝟐𝐱𝟐 + 𝟑𝐱 + 𝟒

𝐱𝟑 + 𝟒𝐱𝟐 + 𝟔𝐱 + 𝟒𝐝𝐱

= ∫ (2

x + 2−

1

x2 + 2x + 2 )

= 2 ∫1

x + 2dx − ∫

1

x2 + 2x + 2dx

= 2 ∫1

x + 2dx − ∫

1

(x + 1)2 + 1dx

u = x + 1 du = dx

= 2 ∫1

x + 2dx − ∫

1

u2 + 1du

=1

x + 2= s = x + 2 ds = dx

= 2 ∫1

sds − tan−1u

= 2 log(s) − tan−1(x + 1) + C

= 𝟐 𝒍𝒐𝒈 (𝒙 + 𝟐) − 𝒕𝒂𝒏−𝟏(𝒙 + 𝟏) + 𝑪