Tugas Matematika

Integral Hal 49- 59

Disusun Oleh :

POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG

TAHUN AJARAN 2014/2015

Industri Air Kantung Sungailiat 33211

Bangka Induk, Propinsi Kepulauan Bangka Belitung

Telp : +62717 93586

Fax : +6271793585 email : polman@polman-babel.ac.id

http://www.polman-babel.ac.id

Kelompok 7 :

- Fery Ardiansyah

- Sarman

- Rakam Tiano

- Mirza ramadhan

Dua aturan integrasi berguna

Latihan 7.7

Cari integral tak tentu yang paling umum..

1. 3𝑥4 − 5𝑥3 − 21𝑥2 + 36𝑥 − 10 𝑑𝑥

2. 3𝑥2 − 4𝑐𝑜𝑠 2𝑥 𝑑𝑥

3. 8

𝑡5+

5

𝑡 𝑑𝑡

4. 1

25 − 𝜃2+

1

100 + 𝜃2 𝑑𝜃

5. 𝑒5𝑥 − 𝑒4𝑥

𝑒2𝑥𝑑𝑥

6. 𝑥7 + 𝑥4

𝑥5 𝑑𝑥

7. 𝑥7 + 𝑥4

𝑥5 𝑑𝑥

8. 𝑥2 + 4 2𝑑𝑥 = 𝑥4

9. 7

𝑡3 𝑑𝑡

10. 20 + 𝑥

𝑥𝑑𝑥

Penyelesaian :

1. 3𝑥4 − 5𝑥3 − 21𝑥2 + 36𝑥 − 10 𝑑𝑥 = 3𝑥4 𝑑𝑥 − 5𝑥3 𝑑𝑥 − 21𝑥2 𝑑𝑥 +

36𝑥 𝑑𝑥 − 10 𝑑𝑥 = 3 𝑥4 𝑑𝑥 − 5 𝑥3 𝑑𝑥 − 21 𝑥2 𝑑𝑥 + 36 𝑥 𝑑𝑥 −

10 𝑑𝑥 = 3 𝑥5

5 − 5

𝑥4

4 − 21

𝑥3

3 + 36

𝑥2

2 − 10𝑥 + 𝑐 =

3

5𝑥5 −

5

4𝑥4 − 7𝑥3 +

18𝑥2 − 10𝑥 + 𝑐

2. 3𝑥2 − 4𝑐𝑜𝑠 2𝑥 𝑑𝑥 = 3𝑥2 𝑑𝑥 − 4 𝑐𝑜𝑠 2𝑥 𝑑𝑥 = 3 𝑥2 𝑑𝑥 − 4 𝑐𝑜𝑠 2𝑥 𝑑𝑥 =

3 𝑥3

3 − 4

1

2𝑠𝑖𝑛2𝑥 + 𝑐 = 𝑥3 − 2 sin 2𝑥 + 𝑐

3. 8

𝑡5 +5

𝑡 𝑑𝑡 =

8

𝑡5 𝑑𝑥 + 5

𝑡𝑑𝑥 = 8 𝑡−5 𝑑𝑥 + 5

1

𝑡𝑑𝑥 = 8

𝑡−4

−4+ 5 𝑙𝑛 𝑡 + 𝑐 =

−2𝑡−4 + 5 𝑙𝑛 𝑡 + 𝑐

4. 1

25−𝜃2+

1

100 +𝜃2 𝑑𝜃 = 1

25−𝜃2𝑑𝑥 +

1

100+𝜃2 𝑑𝑥 = 1

52+𝜃2𝑑𝑥 +

1

102 +𝜃2 𝑑𝑥 =

𝑠𝑖𝑛−1 𝜃

5 +

1

10𝑡𝑎𝑛−1 𝜃

10+ 𝑐

5. 𝑒5𝑥−𝑒4𝑥

𝑒2𝑥 𝑑𝑥 = 𝑒3𝑥 − 𝑒2𝑥 𝑑𝑥 = 𝑒3𝑥𝑑𝑥 − 𝑒2𝑥𝑑𝑥 =1

3𝑒3𝑥 −

1

2𝑒2𝑥 + 𝑐

6. 𝑥7+𝑥4

𝑥5 𝑑𝑥 = 𝑥7

𝑥5 𝑑𝑥 + 𝑥4

𝑥5 𝑑𝑥 =

7. 1

𝑒6+𝑥2 𝑑𝑥 = 𝑒6 + 𝑥2 𝑑𝑥 = 𝑙𝑛 𝑒6 + 𝑥2 + 𝑐

8. 𝑥2 + 4 2𝑑𝑥 = 𝑥4 + 16 + 2.𝑥2. 4 𝑑𝑥 = 𝑥4 + 8𝑥2 + 16 𝑑𝑥 =1

4+1𝑥4+1 +

8

2+1𝑥2+1 + 16𝑥 + 𝑐 =

1

5𝑥5 +

8

3𝑥3 + 𝑐

9. 7

𝑡3 𝑑𝑡 = 7𝑡−

13 𝑑𝑡 =

7

−1

3+1𝑡−

1

3+1 + 𝑐 =

72

3 𝑡

23 + 𝑐 =

21

2𝑡

23 + 𝑐

10. 20+𝑥

𝑥𝑑𝑥 = 20 + 𝑥 𝑥−

12 𝑑𝑥 = 20𝑥−

12 + 𝑥

12 𝑑𝑥 =

20

−12

+1𝑥−

12

+1 +1

12

+1𝑥

12

+1 + 𝑐 =

2012

𝑥12 +

13

2 𝑥

32 + 𝑐 = 40𝑥

12 +

2

3𝑥

32 + 𝑐

Integrasi dasar teknik Integrasi dengan substitusi

Latihan 8.1

Gunakan integrasi dengan substitusi untuk menemukan integral tak tentu yang paling umum.

1. 3 𝑥3 − 5 4𝑥2 𝑑𝑥

2. 𝑒𝑥4𝑥3 𝑑𝑥

3. 𝑡

𝑡2 + 7 𝑑𝑡

4. 𝑥5 − 3𝑥 14 5𝑥4 − 3 𝑑𝑥

5. 𝑥3 − 2𝑥

𝑥4 − 4𝑥2 + 5 4 𝑑𝑥

6. 𝑥3 − 2𝑥

𝑥4 − 4𝑥2 + 5𝑑𝑥

7. cos 3𝑥2 + 1 𝑑𝑥

8. 3𝑐𝑜𝑠2 𝑥(𝑠𝑖𝑛 𝑥)

𝑥 𝑑𝑥

9. 𝑒2𝑥

1 + 𝑒4𝑥 𝑑𝑥

10. 6𝑡2𝑒𝑡3−2 𝑑𝑡

PENYELESAIAN 1. 3 𝑥3 − 5 4𝑥2 𝑑𝑥

u = x3 – 5 du = 3x2 dx

= 𝑢4 𝑑𝑢

=1

5𝑢5 + 𝑐

=(𝑥3 − 5)5

5+ 𝑐

2. 𝑒𝑥4𝑥3 𝑑𝑥

𝑢 = 𝑥4

= 𝑒𝑥4 1

4. 4𝑥3 𝑑𝑥

=1

4 𝑒𝑥

34𝑥3 𝑑𝑥

=1

4 𝑒𝑢 𝑑𝑢

=1

4𝑒𝑢 + 𝑐

=1

4𝑒𝑥

4+ 𝑐

3. 𝑡

𝑡2 + 7 𝑑𝑡

𝑢 = 𝑡2 + 7 𝑑𝑢 = 2𝑡 𝑑𝑥

𝑡

𝑡2 + 7𝑑𝑡

1

2

2𝑡

𝑡2 + 7𝑑𝑡

1

2

2𝑡

𝑡2 + 7𝑑𝑡

1

2 𝑑𝑢

𝑢

1

2𝐼𝑛 𝑢 + 𝑐

1

2𝐼𝑛 𝑡2 + 7 + 𝑐

4. 𝑥5 − 3𝑥 14 5𝑥4 − 3 𝑑𝑥

𝑢 = 𝑥5 − 3𝑥 𝑑𝑢 = 5𝑥4 − 3 𝑑𝑥

= 𝑢14 𝑑𝑢

= 4𝑢54 + 𝑐

= 4 𝑥5 − 3𝑥 54 + 𝑐

5. 𝑥3 − 2𝑥

𝑥4 − 4𝑥2 + 5 4 𝑑𝑥

𝑢 = 𝑥4 − 4𝑥2 + 5 𝑑𝑢 = 4𝑥3 − 8𝑥 𝑑𝑥

= 1

4.4 𝑥3 − 2𝑥

𝑢4 𝑑𝑥

= 1

4 𝑑𝑢

𝑢4

=1

4𝐼𝑛 𝑢 + 𝑐

=1

4𝐼𝑛 𝑥4 − 4𝑥2 + 5 + 𝑐

6. 𝑥3 − 2𝑥

𝑥4 − 4𝑥2 + 5𝑑𝑥

𝑢 = 𝑥4 − 4𝑥2 + 5 𝑑𝑢 = 4𝑥3 − 8𝑥 𝑑𝑥

= 4 𝑥3 − 2𝑥

= 1

4.

4(𝑥3 − 2𝑥)

𝑥4 − 4𝑥2 + 5 𝑑𝑥

=1

4 𝑑𝑢

𝑢

=1

4𝐼𝑛 𝑢 + 𝑐

=1

4𝐼𝑛 𝑥4 − 4𝑥2 + 5 + 𝑐

9. 𝑒2𝑥

1 + 𝑒4𝑥 𝑑𝑥

= 𝑒2𝑥

1 + 𝑒2𝑥(2) 𝑑𝑥

𝑢 = 1 + 𝑒2𝑥 𝑑𝑢 = 2. 𝑒2𝑥 𝑑𝑥

= 1

2.

2. 𝑒2𝑥

1 + 𝑒2𝑥(2)

=1

2 𝑑𝑢

𝑢

=1

2𝐼𝑛 𝑢 𝑑𝑥

=1

2𝐼𝑛 1 + 𝑒4𝑥 + 𝑐

10. 6𝑡2𝑒𝑡3−2 𝑑𝑡

𝑢 = 𝑡3 − 2 𝑑𝑢 = 3𝑡2 𝑑𝑡

= 6𝑡2𝑒𝑡3−2 𝑑𝑡

= 2 3𝑡2 𝑒𝑡3−2 𝑑𝑡

= 1

3. 3 2 . 3𝑡2 . 𝑒𝑡

3−2 𝑑𝑡

=1

3 6 𝑑𝑢. 𝑒𝑢

=1

3𝑒𝑢 . 6 𝑑𝑢

=1

3𝑒𝑡

3−2. 6 + 𝑐

= 2𝑒𝑡3−2 + 𝑐

Integrasi dengan bagian

Latihan 8.2

Gunakan integrasi dengan bagian untuk menemukan integral tak tentu yang paling umum.

1. 2𝑥.sin2x dx

2. 𝑥3lnx dx

3. 𝑡𝑒𝑡dt

4. 𝑥 cos x dx

5. 𝑐𝑜𝑡−1 𝑥 𝑑𝑥

6. 𝑥2 𝑒𝑥𝑑𝑥

7. 𝑤( 𝑤 − 3)2 𝑑𝑤

8. 𝑥3 𝑖𝑛 4𝑥 𝑑𝑥

9. 𝑡 (𝑡 + 5)−4 𝑑𝑡

10. 𝑥 𝑥 + 2 .𝑑𝑥

PENYELESAIAN

1. 2𝑥 sin 2𝑥 𝑑𝑥

Misalnya :

u = 2x du = x

dv = sin 2x dx v= sin 2𝑥𝑑𝑥 = - 1

2 cos2x

𝑢.𝑑𝑣 = 𝑢𝑣 – 𝑢. 𝑑𝑢

2𝑥 sin 2𝑥 𝑑𝑥 = (2x) (- 1

2 cos 2x ) - (−

1

2 cos 2x ) . 2x

= - 2

2 cos 2x +

1

2 cos 2x dx

= - x cos 2x + 1

2 .

1

2 sin 2x

= - x cos 2x + 1

2 . sin 2x + c

2. 𝑥3 𝑖𝑛 𝑥 𝑑𝑥

Misalnya :

U= inx du = 1

𝑥 dx

dv= 𝑥3dx v = 𝑥3 𝑑𝑥 = 𝑥4

4

𝑢.𝑑𝑣 = 𝑢𝑣 – 𝑢. 𝑑𝑢

𝑥3 𝑖𝑛 𝑥 𝑑𝑥 = (in x) (𝑥4

4) -

𝑥4

4 .

1

𝑥 dx

= 𝑥4𝑖𝑛𝑥

4 -

1

4 . 𝑥4

4

= 𝑥4𝑖𝑛𝑥

4 -

𝑥4

16 + c

3. 𝑡𝑒𝑡 𝑑𝑡

Misalnya :

U = t du = dt

dv = 𝑒𝑡 dt v = 𝑒𝑡dt = 𝑒𝑡

𝑢.𝑑𝑣 = 𝑢. 𝑣 – 𝑢. 𝑑𝑢

𝑡𝑒𝑡 𝑑𝑡 = (t) (𝑒𝑡) - 𝑒𝑡dt

= 𝑡𝑒𝑡 - 𝑒𝑡dt

= 𝑡𝑒𝑡 - 𝑒𝑡 + c

4. 𝑥 cos 𝑥 𝑑𝑥

Misalnya :

U= x du = dx

dv = cos x dx v = cos 𝑥 𝑑𝑥 = sin x

𝑢.𝑑𝑣 = 𝑢. 𝑣 – 𝑢. 𝑑𝑢

𝑥 cos 𝑥 𝑑𝑥 = ( x ) ( sin x ) - sin 𝑥 𝑑𝑥

= sin x + cosx dx

= sin x + cosx + c

5. 𝑐𝑜𝑡−1( x ) dx

Misalnya :

U = sin𝑥−1

Du= cos𝑥−1

Subtitusi du = sin𝑥−1 du = cos𝑥−1

𝑐𝑜𝑠𝑥 −1

𝑠𝑖𝑛𝑥 −1 dx = 𝑑𝑢

𝑢

Salve integral

= in (u) + c

Subsitusi kembali

U=sin𝑥−1

= in (sin𝑥−1) + 𝑐

6. 𝑥2 𝑒𝑥𝑑𝑥

Misalnya :

U = 𝑥2 du = 2x

dv = 𝑒𝑥dx v = 𝑒𝑥dx = 𝑒𝑥

𝑢.𝑑𝑣 = u.v - 𝑢.du

𝑥2 𝑒𝑥𝑑𝑥 = 𝑥2𝑒𝑥- 𝑥2

. 2𝑥

=𝑥𝑒2𝑥 - 2𝑥. 𝑑𝑥

=𝑥𝑒2𝑥 - x+c

7. 𝑤(𝑤 − 3)2𝑑𝑤

Misalnya :

U= w du= dw

dv = (𝑤 − 3)2𝑑𝑤 𝑣 = 2𝑤 − 6 = 𝑤 − 3

𝑢.𝑑𝑣 = u.v - 𝑢.du

𝑤(𝑤 − 3)2𝑑𝑤 = 𝑤. 𝑤 − 3 − 𝑤. 𝑑𝑤

= 𝑤2 − 3𝑤 −1

2𝑤 + 𝑐

8. 𝑥3 𝑖𝑛 4𝑥 𝑑𝑥

Misalnya :

U= in4x du= 1

4𝑥𝑑𝑥

dv= 𝑥3𝑑𝑥 v = 𝑥3dx = 1

4𝑥4

𝑢.𝑑𝑣 = u.v - 𝑣.du

𝑥3 𝑖𝑛 4𝑥 𝑑𝑥 = in4x. 1

4𝑥4- in4x .

1

4𝑥𝑑𝑥

= 1

4𝑥4𝑖𝑛4𝑥 −

1

5𝑥5 ∶

1

2 16𝑥2 + 𝑐

= 1

4𝑥4𝑖𝑛4𝑥 -

2𝑥5

80𝑥2 + c

9. 𝑡(𝑡 + 5)−4 𝑑𝑡

Misalnya :

U= t du= dt

dv =(𝑡 + 5)−4 𝑣 = −4𝑡−3 − 20−3 = 2𝑡−2 + 10−2

𝑢.𝑑𝑣 = u.v - 𝑣.du

𝑡(𝑡 + 5)−4 𝑑𝑡 =( t. 2𝑡−2 + 10−2 ) - 2𝑡−2 + 10−2 .𝑑𝑡

= 20𝑡−4 + (2𝑡 + 10 + 𝑑𝑡

10. 𝑥 𝑥 + 2 .dx

Misalnya :

U = x du = dx

Dv= 𝑥 + 2 dx v= (𝑥 + 2)1

2 =2𝑥11

2 +0.6711

2

𝑢.𝑑𝑣 = u.v - 𝑣.du

𝑥 𝑥 + 2 .dx = x . 2𝑥11

2 +0.6711

2 - 2𝑥11

2 + 0.6711

2 . dx

= x.2,67𝑥3

2 - (2𝑥3

2 + 0,673

2) dx

= 2,67𝑥23

2 - 2,67𝑥6

2 + c

Integrasi dengan menggunakan tabel rumus

terpisahkan

Latihan 8.3

Gunakan tabel rumus integral dalam Lampiran C untuk menemukan integral tak tentu yang paling umum.

1. cot 𝑥 𝑑𝑥

2. 1

𝑥+2 (2𝑥+5)𝑑𝑥

3. 𝑙𝑛𝑥 2𝑑𝑥

4. 𝑥 cos𝑥 𝑑𝑥

5. 𝑥

𝑥+2 2 𝑑𝑥

6. 3𝑥𝑒𝑥 𝑑𝑥

7. 10 𝑤 + 3 𝑑𝑤

8. 𝑡(𝑡 + 5)−1 𝑑𝑡

9. 𝑥 𝑥 + 2 𝑑𝑥

10. 1

sin𝑢 cos𝑢 𝑑𝑢

PENYELESAIAN

1. cot 𝑥 𝑑𝑥 ( Formula nomor 7) Penyelesaian :

𝑐𝑜𝑡 𝑥 𝑑𝑥 = 𝑐𝑜𝑠𝑥

𝑠𝑖𝑛𝑥 𝑑𝑥

Misalkan : 𝑢 = sin 𝑥 𝑑𝑢 = cos𝑥 𝑑𝑥 Subsitusi 𝑑𝑢 = cos 𝑥,𝑈 = sin 𝑥

cos𝑥

sin 𝑥 𝑑𝑥 =

𝑑𝑢

𝑢

𝑠𝑎𝑙𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 ln 𝑢 + 𝐶 subsitusi kembali 𝑈 = sin 𝑥 𝑙𝑛 sin 𝑥 + 𝑐

2. 1

𝑥+2 (2𝑥+5)𝑑𝑥

=1

𝑥 + 2 (2𝑥 + 5)=

𝐴

𝑥 + 2+

𝐴

2𝑥 + 5

𝐴 =1

𝑥 + 2 (2.2 + 5)=

1

9

𝐵 =1

5 + 2 (2𝑥 + 5)=

1

7

Sehingga :

1

𝑥 + 2 2𝑥 + 5 𝑑𝑥 =

1

𝑥 + 2 2𝑥 + 5

=

19

𝑥 + 2 𝑑𝑥 +

19

2𝑥 + 5 𝑑𝑥

=1

9𝑙𝑛 𝑥 + 2 +

1

7ln 2𝑥 + 5 + c

3. 𝑙𝑛𝑥 2𝑑𝑥 = 𝑙𝑛𝑥 𝑙𝑛𝑥 𝑑𝑥 Missal :

U = ln x 𝑑𝑢 = (1

𝑥 )2

Dv = dx dv = 𝑑𝑥 v = x

(𝑙𝑛𝑥)2 𝑑𝑥 = 𝑢𝑣 − 𝑣𝑑𝑢 (x ln )

= (𝑙𝑛𝑥)2. x - 𝑥 1

𝑥2 𝑑𝑥

= 𝑥. (𝑙𝑛𝑥)2 - 1

𝑥 𝑑𝑥

= 𝑥. (𝑙𝑛𝑥)2 x - 𝑥−1 𝑑𝑥

= 𝑥. (𝑙𝑛𝑥)2 - 1

0𝑥0 + 𝑐

= 𝑥. (𝑙𝑛𝑥)2 - ~ + 𝑐

= ln x ( x ln x-x ) – (𝑥 ln 𝑥 − 𝑥) . 1

𝑥

=x (ln x)2 - x ln x -

4. 𝑥 cos𝑥 𝑑𝑥

Penyelesaian : 𝑈 = 𝑋 → 𝑑𝑢 = 𝑑𝑥

𝑑𝑣 = 𝑐𝑜𝑠𝑥 → 𝑣 = 𝑠𝑖𝑛𝑥

𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢

𝑥𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥𝑠𝑖𝑛𝑥 − 𝑠𝑖𝑛𝑥 𝑑𝑥

𝑥𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝑐

5. 𝑥

𝑥+2 2 𝑑𝑥

Penyelesaian :

𝑥

𝑥+2 2 =𝐴

𝑥+2 +𝐵

𝑥+2 =𝐴 𝑥+2 +𝐵

𝑥+2 2

𝐴 = 2

𝐴 + 𝐵 = 0 = −2

Sehingga :

𝑥

𝑥 + 2 2 𝑑𝑥 =

𝑑𝑥

𝑥 + 2 –

𝑑𝑥

𝑥 + 2 2

𝑀𝑖𝑠𝑎𝑙𝑙 𝑢 = 𝑥 + 2 → 𝑑𝑢 = 𝑑𝑥

𝑑𝑥

𝑥 + 2 –

𝑑𝑥

𝑥 + 2 2 = 𝑑𝑢

𝑢 –

𝑑𝑢

𝑢2= 2𝑙𝑛 +

2

𝑢+ 𝑐

2𝑙𝑛 𝑥 + 2 +2

𝑥+2 + 𝑐

6. 3𝑥𝑒𝑥 𝑑𝑥

U = 3x dv = 𝑒𝑥 𝑑𝑥 𝑑𝑢

𝑑𝑥= 3 v = 𝑒𝑥 𝑑𝑥 = 𝑒𝑥

du = 3 dx

𝑢𝑑𝑣 = u.v – 𝑣 𝑑𝑢

= (3x) . (𝑒𝑥) – 𝑒𝑥 . 3 𝑑𝑥 = 3x 𝑒𝑥 − 3𝑒𝑥

7. 10 𝑤 + 3 dw ( Formula nomor 2)

10 𝑤 + 3 dw = (10 𝑤 + 3)1

2 dw

=1

12 + 1

(10 𝑤 + 3)12

+1 + 𝑐

=2

3 (10 𝑤 + 3)

3

2 + 𝑐

8. 𝑡(𝑡 + 5)−1 𝑑𝑡

= 𝑡

𝑡+5 dt = 𝑡 (𝑡 + 5)−1 𝑑𝑡

Missal:

U = t + 5 U= t+5

𝑑𝑢

𝑑𝑡 = 1 t = (u-5)

𝑑𝑢 = 𝑑𝑡 t=u→u=t+5 =5

t = 2 → u=t+5 = 7

= 𝑡

𝑡+5 dt = 𝑡 (𝑡 + 5)−1 𝑑𝑡 = 𝑢 − 5 𝑢−1 𝑑𝑢 = 𝑢0 − 5𝑢−1 𝑑𝑢

(𝑢0 − 5𝑢)………… . = 𝑢 − 𝑢

−5𝑢−1 +1 du

−5(𝑢1 −1

5 𝑥 ) 𝑑𝑥

-5 (ln 𝑢 - 15

0+1

𝑥0+1)

-5 ( ln 𝑡 + 5 - 1

5 x)

-5 ln 𝑡 + 5 + x

9. 𝑥 𝑥 + 2 𝑑𝑥 𝑚𝑖𝑠𝑎𝑙 𝑢 = 𝑥 + 2 → 𝑥 = 𝑢 − 2 𝑑𝑢 = 𝑑𝑥

Sehingga integral diatas dapat menjadi :

= 𝑖𝑛𝑡 𝑢 − 2 𝑈 𝑑𝑢

= 𝑖𝑛𝑡 𝑢 − 2 𝑈1

2 𝑑𝑢

= 𝑖𝑛𝑡 𝑈5

2 − 𝑈1

2 𝑑𝑢

=2

7 𝑈

27 −

2

3𝑈

32 + 𝐶

= 𝑖𝑛𝑡 (𝑥 + 2)52 −

2

3 (𝑥 + 2)

3 2 + 𝐶