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)] Therefore,
2
12
1
C
1
(
1)
x
x
x
x
e dx
e
x
x
+
−
=
+
+
+
∫
EXERCISE 7 6
Integrate the functions in Exercises 1 to 22 1