1 . (𝑥10 −6

𝑥5+ 𝑥73

) 𝑑𝑥 = (𝑥10 − 6𝑥−5 + 𝑥7

3) 𝑑𝑥

=1

11𝑥11 +

6

4𝑥−4 +

3

10𝑥

10

3 +c

=1

11𝑥11 +

3

2𝑥−4 +

3

10𝑥

10

3 +c

2. cos 9𝑥 − 11 + 𝑠𝑒𝑐2 6𝑥 − 8 𝑑𝑥

=1

9sin 9𝑥 − 11 +

1

6tan 6𝑥 − 8 + 𝑐

3. Dengan menggunakan cara subsitusi

𝑥

6+𝑥2 𝑑𝑥 = 𝑥(6 + 𝑥2)

1

2 𝑑𝑥

Misalkan :

𝑢 = 6 + 𝑥2

𝑑𝑢

𝑑𝑥= 2𝑥

𝑑𝑢 =1

2𝑥 𝑑𝑢

𝑥(6 + 𝑥2)1

2 𝑑𝑥

= 𝑥 𝑈−1

2 .1

2𝑥𝑑𝑢

= 𝑥

2𝑥.𝑈

−1

2 𝑑𝑢

= 1

2.𝑈

−1

2 𝑑𝑢

=1

2−1

2+1

𝑈−1

2+1 + 𝐶

=1

21

2

𝑈1

2 + 𝐶

=(6𝑥 + 𝑥2)1

2 + 𝐶

4. Dengan menggunkan cara subsitusi

2𝑥 + 5 cos(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥

Misalkan

U =2𝑥2 + 10𝑥 + 8

𝑑𝑢

𝑑𝑥= 4𝑥 + 10

Dx=1

4𝑥+10 𝑑𝑢

2𝑥 + 5 cos(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥

= 2𝑥 + 5 cos 𝑢 1

4𝑥+10 du

2𝑥 + 5

2 2𝑥 + 5 cos 𝑢 𝑑𝑢

1

2 cos u du

=1

2 sin u du

=1

2 sin (2𝑥 2+ 10x +8 ) + c

5.Integral parsial

2𝑥. sin(10𝑥 + 3) dx

Misalkan :

u= 2x du =2dx

dv =sin (10x +3 ) v= sin 10𝑥 + 3 𝑑𝑥 = −1

10cos 10𝑥 + 3

= 𝑈𝑑𝑣 = 𝑢𝑣 − 𝑣 𝑑𝑢

= 2𝑥. sin 10𝑥 + 3 𝑑𝑥

=2𝑥 −1

10cos 10𝑥 + 3 − −

1

10cos 10𝑥 + 3 . 2 𝑑𝑥

=−1

5𝑥. cos 10𝑥 + 3 𝑑𝑥 +

2

100sin 10𝑥 + 3 + 𝐶

=−1

5𝑥. cos 10𝑥 + 3 +

1

50sin 10𝑥 + 3 + 𝐶

6. Dengan menggunakan table

𝑥2𝑒−7𝑥 𝑑𝑥

Turunan U Integral dv

+𝑥2 -2x +2 -0

𝑒−7𝑥

−1

7𝑒−7𝑥

1

49𝑒−7𝑥

−1

363𝑒−7𝑥

𝑢𝑑𝑣 = 𝑥2 ( −1

7𝑒−7𝑥) -2x .

1

49𝑒−7𝑥 + 2 −

1

369𝑒−7𝑥 + 𝑐

= −𝑥2 1

7𝑒−7𝑥 -2x .

1

49 −

2

363𝑒−7𝑥𝑒−7𝑥 + 2 + 𝑐

= −1

7𝑥2𝑒−7𝑥 -2x .

1

49 −

2

363𝑒−7𝑥𝑒−7𝑥 + 2 + 𝑐

7.Integral fungsi rasional

𝑥

𝑥2 − 2𝑥 − 35 𝑑𝑥

𝑥

𝑥2−2𝑥−35 =

𝑥

𝑥−7 (𝑥+5)=

𝐴

(𝑥−7)+

𝐵

(𝑥+5)

=𝐴 𝑥 + 5 + 𝐵(𝑥 − 7)

𝑥 − 7 𝑥 + 5

𝐴𝑥 + 5𝐴 + 𝐵𝑥 − 7𝐵)

𝑥 − 7 𝑥 + 5

A+B = 1 x5 5A+5B = 5

5A +B =0 x1 5A-7B = 0

12B=5

B=5

12 A=

7

12

Sehingga :

𝑥

𝑥 − 7 (𝑥 + 5) 𝑑𝑥 =

𝐴

𝑥 − 7 𝑑𝑥 +

𝐵

𝑥 + 5 𝑑𝑥

= 7

12

𝑥−7 𝑑𝑥 +

5

12

𝑥+5 𝑑𝑥

= 7

12 𝑙𝑛 x-7 +

5

12𝑙𝑛 x+5 + C

8. ( 𝑥45

1+ 3𝑥 +

1

𝑥3 ) 𝑑𝑥 = ( 𝑥45

1+ 3𝑥 + 𝑥−3 ) 𝑑𝑥

= 1

5[𝑥5 +

3

2𝑥2 −

1

2𝑥−2] 5

1

= (1

555 +

3

252 −

1

2𝑥5−2 ) –(

1

515 +

3

212 −

1

21−2 )

=(625 + 75

2−

1

50 ) − (

1

5+

3

2−

1

2 )

=625- 1 + 75

2−

1

50−

1

5

=624 + 75

2 -

1

50 -

1

5

31200 + 1875 − 1 − 10

50=

33064

50= 661

14

50

9.Dik = y = 𝑥2 − 1

Y = 3x + 9

Dit = Luas daerah

Jawab :

𝑥2 − 1 = 3𝑥 + 9

𝑥2 − 1 − 3𝑥 − 9 = 0

𝑥2 − 3𝑥 − 10 = 0

(x-5) (x+2) = 0

X= 5 v x=-2

L= 3𝑥 + 9 – (𝑥25

−2− 1) 𝑑𝑥

= 3𝑥 −5

−2𝑥2 + 10 𝑑𝑥

=3

2[𝑥2 −

1

3𝑥3 + 10𝑥] 5

−2

= 3

252 −

1

353 + 10.5 −

3

2(−2)2 −

1

3(−2)3 + 10.−2

= 75

2 −

125

3 + 50 − 6 +

8

3− 20

= 225−250+300

6 −

18+8−60

3

=275

6+

34

3 =

275+68

6=

343

6= 57

1

6

10.

Diketahu :

iy = 3x Y= x

Y= 0 y= 2

Dit : Volume benda = mengelilingi sumbu y

Jawab =

V = 𝜋 𝑥2 − 𝑥22 𝑑𝑦 𝑑

𝑒

= 𝜋 (𝑦2 − ( 2

0

1

3𝑦)2 ) dy

= 𝜋 𝑦2 −2

0

1

9𝑦2 dy

= 𝜋 −2

0

8

9𝑦2 dy

=𝜋[(8

9

2+1𝑦2+1)] 2

0 = 𝜋(

8

27𝑦3) 2

0

= 𝜋(8

2723-)-(

8

27. 03) = 𝜋

64

27 =2

10

27 𝜋