(iii)
We know that
1
1
1
(log )
0 and
[log (
)]
(
1)
0
d
d
x
,x
– x
–
,x
dx
x
dx
– x
x
=
>
=
=
<
Combining above, we get 
(
)
1
log
0
d
x
, x
dx
=x
≠
Therefore, 
1
log
dx
x
x
=
∫
 is one of the anti derivatives of 1
x Example 2 Find the following integrals:
(i)
3
2
1
x – dx
∫x
(ii)   
(32
1)
x
dx
+
∫
(iii)   ∫
23
1
(
2
–
)
+
∫
x
x
e
dx
x
Solution
(i)
We have
3
2
2
1
–
x –
dx
x dx –
x
dx
x
=
∫
∫
∫
(by Property V)
296
MATHEMATICS
= 
1
1
2
1
1
2
C
C
1 1
2
1
–
x
x
–
–
+
+




+
+




+
+




;  C1, C2 are constants of integration
= 
2
1
1
2
C
C
2
1
–
x
–x
–
–
+
 = 
2
1
2
1 + C
C
x2
–
x
+
= 
2
1 + C
2
x
x
+
, where C = C 1 – C2 is another constant of integration �Note  From now onwards, we shall write only one constant of integration in the
final answer (ii) We have
2
2
3
3
(
1)
x
dx
x dx
dx
+
=
+
∫
∫
∫
=
2
1
3
C
2
1
3
x
x
+
+
+
+
 = 
335
C
5
x
+x
+
(iii) We have 
3
3
2
2
1
1
(
2
)
2
x
x
x
e –
dx
x
dx
e dx –
dx
x
x
+
=
+
∫
∫
∫
∫
=
3
1
2
2
– log
+ C
3
1
2
x
x
e
x
+
+
+
=
25
2
2
– log
+ C
5
x
x
e
x
+
Example 3 Find the following integrals:
(i)
(sin
cos )
x
x dx
+
∫
(ii) cosec
(cosec 
cot
)
x
x
x dx
+
∫
(iii)
2
1
–cossin
xx dx
∫
Solution
(i)
We have
(sin
cos )
sin
cos
x
x dx
x dx
x dx
+
=
+
∫
∫
∫
= – cos
sin
C
x
x
+
+
INTEGRALS         297
(ii)
We have
2
(cosec
(cosec
+ cot
)
cosec
cosec
cot
x
x
x dx
x dx
x
x dx
=
+
∫
∫
∫
= – cot
cosec
C
x –
x +
(iii)
We have
2
2
2
1 sin
1
sin
cos
cos
cos
–
x
x
dx
dx –
dx
x
x
x
=
∫
∫
∫
= 
sec2
tan
sec
x dx –
x
x dx
∫
∫
= tan
sec
C
x –
x +
Example 4 Find the anti derivative F of  f defined by f (x) = 4x3 – 6, where F (0) = 3
Solution One anti derivative of f (x) is x4 – 6x since
(4
6 )
d
x – x
dx
 = 4x3 – 6
Therefore, the anti derivative F is given by
F(x) = x4 – 6x + C, where C is constant