
Ingat Aturan Rantai pada Turunan :
d
f ( g ( x))  f ' ( g ( x)) g ' ( x)
dx

Jika kedua ruas diintegralkan, maka
diperoleh

d
f ( g ( x)) dx   f ' ( g ( x)) g ' ( x)dx
dx
f ( g ( x))  C   f ' ( g ( x)) g ' ( x)dx
dari definisi integral tak tentu

f ' ( g ( x)) g ' ( x)dx  f ( g ( x))  C

Misal u = g(x), maka du = g’(x)dx

Disubstitusi ke atas diperoleh

f ' (u )du  f (u )  C
1.
2.
3.
4.
5.
6.
Mulai dengan fungsi yang diintegralkan
Kita misalkan u = g(x)
Hitung du
Substitusi u dan du
Integralkan
Ganti u dengan g(x)
Hitunglah

sin( 3 x  5) dx
Jawab
Misalkan u = 3x + 5 , maka du = 3 dx , dx =
1/3 du
Substitusi ke fungsi di atas diperoleh

sin( 3x  5)dx   sin udu   cos u  C   cos( 3 x  5)  C



Hitunglah

9 xe
3 x 2  5
dx
Jawab
Misalkan u = -3x2 + 5 , maka du = -6x dx
atau
x dx = -1/6 du
9 u
9 u
9 3 x  5
3 x 2  5
 9 xe dx    e du   e  C   e  C
2
6
6
6

Hitunglah

Jawab


tan xdx
sin x
tan xdx  
dx
cos x
Misalkan u = cos x , maka du = -sin x dx
atau
sin x dx = -du.
Sehingga


sin x
 du
tan xdx  
dx  
  ln u  C   ln cos x  C  ln sec x  C
cos x
u


Exercise


Bentuk integral
 f ( x) g ( x)dx
dapat
diselesaikan dengan metode Integral By
Parts (Integral sebagian – sebagian) , yaitu
 f ( x) g ' ( x)dx  f ( x) g ( x)   g ( x) f ' ( x)dx
Atau lebih dikenal dengan rumus
 udv  uv   vdu


Hitunglah (3  5 x ) cos( 4 x ) dx

Jawab
Misalkan u = 3 – 5x , du = -5 dx.
dv = cos 4x , v = ¼ sin 4x dx

Maka

1
1
(
3

5
x
)
cos(
4
x
)
dx

(
3

5
x
)(
sin(
4
x
)

4

 4 sin( 4 x)( 5dx )

Hitunglah a
b
c
Exercise
3
(
x
  5) ln( x)dx
2x
e
 cos( x)dx
2
x
 cos( 4 x)dx

Link to James Stewart

The method of Partial Fractions provides a
way to integrate all rational functions.
Recall that a rational function is a function
of the form
P( x)
 Q( x)dx
where P and Q are polynomials.
1. The technique requires that the degree of
the numerator (pembilang) be less than the
degree of the denominator (penyebut)
If this is not the case then we first must
divide the numerator into the denominator.
2. We factor the denominator Q into powers
of distinct linear terms and powers of
distinct quadratic polynomials which do not
have real roots.
3. If r is a real root of order k of Q, then the
partial fraction expansion of P/Q contains a
term of the form

Ak
A1
A2


2
(x  r) (x  r)
( x  r)k
where A1, A2, ..., Ak are unknown constants.
4. If Q has a quadratic factor ax2 + bx + c which
corresponds to a complex root of order k,
then the partial fraction expansion of P/Q
contains a term of the form

Bk x  Ck
B1 x  C1
B2 x  C2




ax2  bx  c (ax2  bx  c) 2
(ax2  bx  c) k
where B1, B2, ..., Bk and C1, C2, ..., Ck are
unknown constants.
5. After determining the partial fraction
expansion of P/Q, we set P/Q equal to the
sum of the terms of the partial fraction
expansion. (See Ex-2.Int.Frac)

6. We then multiply both sides by Q to get
some expression which is equal to P.
7. Now, we use the property that two
polynomials are equal if and only if the
corresponding coefficients are equal.
(see ex3-int.Fractional)
8. We express the integral of P/Q as the sum
of the integrals of the terms of the partial
fraction expansion.
(see Ex4-Int.Fractional)
9. Integrate linear factors:


A1
dx  A1 ln x  r
(x  r)
A1
A1
 n 1
dx

(
x

r
)
for n > 1
( x  r )n
 n 1
10. Integrate quadratic factors:
Some simple formulas:

Bx  C
B
C
 x
2
2
dx  ln(x  a )  arctan 
2
2
x a
2
A
a

Bx  C
Cx  a 2 B
C
 x
dx  2 2
 3 arctan 
2
2 2
2
(x  a )
2a ( x  a ) 2a
a



Hitunglah

x3  6 x 2 1
dx
2
x  4 x  21
Jawab
Link Ex1-Int.Fractional


Exercise
Link to Drii – Int.Fractional

Link to Strategi Pengintegralan

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Evaluate

Answer

Link to Tabel Rumus Umum integral