IB Maths HL
4: Harder Indefinite
Integration
Indefinite Integration
Reminder:

•
•
•
n 1
x
x n dx 
 C;
n1
n  1
add 1 to the power
divide by the new power
add C
n does not need to be an integer BUT notice that the
rule is for x n
It cannot be used directly for terms such as
1
xn
Indefinite Integration
e.g.1 Evaluate
1
 x 4 dx
Solution: Using the law of indices,
So,
1
 x4
This minus sign . . .
. . . makes the
term negative.
dx 

x  4 dx
1
x4
x 3

 C
3
x 3

 C
3
 x4
Indefinite Integration
e.g.1 Evaluate
1
 x 4 dx
Solution: Using the law of indices,
So,
But this one . . .
is an index
1
 x4
dx 

x  4 dx
1
x4
x 3

 C
3
x 3

 C
3
1

 C
3x3
 x4
Indefinite Integration
e.g.2 Evaluate
Solution:

1
x2

1
x2
dx 
dx
3
x2
3
2
 C
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
3
x2
3
2
2
  C 
2
3
2x 2
3
 C
We can get this answer directly by noticing that . . .
. . . dividing by a fraction is the same as multiplying
by its reciprocal. ( We “flip” the fraction over ).
Indefinite Integration
e.g.2 Evaluate
Solution:

1
x2

1
x2
dx 
dx
3
x2
3
2
 C
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
3
x2
3
2
3
x2
2
2
  C 
2
3
 C
We can get this answer directly by noticing that . . .
. . . dividing by a fraction is the same as multiplying
by its reciprocal. ( We “flip” the fraction over ).
Indefinite Integration
e.g.3 Evaluate
Solution:
So,

x x

1
x
dx
1
2
1
x
dx 
Using the law of indices,



1
x
x
x
1
2
 12
1
2
1
2
 2x
dx
dx
 C
1
2
 C
Indefinite Integration
e.g.4 Evaluate
Solution:


x 1
dx
x
x 1
dx

x

x 1
1
x2
dx
1
We cannot Split
integrate
with
x inx the
denominator.
up xthe
Write
infraction
index
form


dx



1
x2
1
x2
Use the 2nd law of indices:
x

x
1
x2
1
2
dx

1 1
x 2

1
x2
Indefinite Integration
e.g.4 Evaluate
Solution:


x 1
dx
x
x 1
dx




x

The terms are now in
our rule

x 1
1
x2
x
1
x2
1
x2

dx
1
x

1
2
1
x 2
3
1
0

1
the form
2 where
21
2
x

x
2
and
2
x
1

of integration.
x32
Instead of
of dividing
dividing by
by
Instead
31
22
dx
dx
1

2
we
x can
C
,multiply by
by 22
,multiply
3
use
Indefinite Integration
e.g.5 The curve y  f ( x ) passes through the point
( 1, 0 ) and f ( x )  x 
/
2
1
x
2
Find the equation of the curve.
1
Solution:
y  x 2  2 dx
x
y  to
x 2 prepare
 x 2 dx all the terms
It’s important

before integrating any of them
x3 1
x 3 x 1
 C

y

C
 y
3 x
3
1


( 1, 0 ) on the curve:
So the curve is
1
2
1  C 
C
3
3
x3
1
2
y


3
x
3

0
Indefinite Integration
Exercise
Evaluate
x
1
1.
Solution:
1
x
3
3
dx 
2.
dx
x
3
dx


x ( x  1) dx

2x
2
C

3
x2
x ( x  1) dx 
x 2

C
2
1

1
x 2 ( x  1) dx


5
2x 2
5

3
2x 2
3

1
x2
C
dx
Indefinite Integration
Exercise
2
dy
x
 1 , find the equation of
3. Given that

dx
x2
the curve through the point ( 2, 0 ).
Solution:
dy x 2  1
2

y

1

x
dx

2
dx
x
x 1
1
 y  x
C  y  x C
1
x
1
3
( 2, 0 ) on the curve:  0  2   C
  C
2
2
So the curve is

1
3
y x 

x
2
Indefinite Integration
Indefinite Integration
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Indefinite Integration
e.g.1 Evaluate
1
 x 4 dx
Solution: Using the law of indices,
So,
1
 x4
This minus sign . . .
. . . makes the
term negative.
dx 

x  4 dx
1
x4
x 3

 C
3
x 3

 C
3
1

 C
3x3
 x4
But this one
is an index
Indefinite Integration
e.g.2 Evaluate
Solution:

1
x2

1
x2
dx 
dx
3
x2
3
2
 C
We need to simplify this “piled up” fraction.
Multiplying the numerator and denominator by 2 gives
3
x2
3
2
2
  C 
2
3
2x 2
3
 C
We can get this answer directly by noticing that . . .
. . . dividing by a fraction is the same as multiplying
by its reciprocal. ( We “flip” the fraction over ).
Indefinite Integration
e.g.3 Evaluate
Solution:
So,

x x

1
x
dx
1
2
1
x
dx 
Using the law of indices,



1
x
x
x
1
2
 12
1
2
1
2
 2x
dx
dx
 C
1
2
 C
Indefinite Integration
x 1
dx
e.g.4 Evaluate
x

Solution:
We cannot integrate with x in the denominator.
Write
x in index form
Split up the fraction
Use the laws of indices:
x
1
x2





x 1
1
x2
x
1
x2
1 1
x 2

dx
1
x

1
2
1
x2
dx
and
1
x
1
2
x
1
2
Indefinite Integration
So,

x
x
1
2

1
x
1
2
dx 

1
x2

1
x 2
dx
The terms are now in the form where we can use
our rule of integration.

3
2x 2
3

1
2x 2
C
Indefinite Integration
e.g.5 The curve y  f ( x ) passes through the point
( 1, 0 ) and f ( x )  x  .
/
2
1
x
2
Find the equation of the curve.
1
Solution:
y  x 2  2 dx
x
y  to
x 2 prepare
 x 2 dx all the terms
It’s important

before integrating any of them


x3 1
x 3 x 1
 C

y

C
 y
3 x
3
1
( 1, 0 ) on the curve:  0  1  1  C  2  C
3
So the curve is
x3
1
2
y


3
x
3
3