6 1 Integral of the type 
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
We have
I =
[
( ) +
( )]
ex
f x
f
x
dx
′
∫
 = 
( )
+
( )
x
x
e f x dx
e f
′x dx
∫
∫
=
1
1
I
( )
, where I =
( )
x
x
e f
x dx
e f x dx
′
+ ∫
∫ (1)
Taking f(x) and ex as the first function and second function, respectively, in I 1 and
integrating it by parts, we have I1 = f (x) ex – 
( )
C
x
f
′x e dx
+
∫
Substituting I1 in (1), we get
I =
( )
( )
( )
C
x
x
x
e f x
f
x e dx
e f
x dx
′
′
−
+
+
∫
∫
 = ex f (x) + C
INTEGRALS         327
Thus,
′
∫
[
( )
( )]
ex
 f x + f
x
dx  = 
( )
C
x
e f x +
Example 22 Find (i) 
1
2
1
(tan
)
1
x
–
e
x
x
+
+
∫
dx   (ii) 
2
2
(
+ 1)
( + 1)
x
x
e
∫x
 dx
Solution
(i) We have I =
1
12
(tan
)
1
x
–
e
x
dx
x
+
+
∫
Consider f (x) = tan– 1x, then  f ′(x) = 
2
1
1
x
+
Thus, the given integrand is of the form ex [ f (x) + f ′(x)] Therefore, 
1
12
I
(tan
)
1
x
–
e
x
dx
x
=
+
+
∫
 = ex tan– 1x + C
(ii) We have 
2
2
(
+1)
I
( +1)
x
x
e
= ∫x
dx
2
2
1 +1+1)
[
]
( +1)
x
x –
e
dx
x
= ∫
2
2
2
1
2
[
]
( +1)
( +1)
x
x –
e
dx
x
x
=
+
∫
 
2
1
2
[
+
]
+1
( +1)
x
x –
e
dx
x
x
= ∫
Consider 
1
( )
1
x
f x
x
−
=
+
, then  
2
2
( )
(
1)
f
x
x
′
=
+
Thus, the given integrand is of the form ex [f (x) + f ′(x)]