
13-26 Calculus – Differentiation and Integration
To evaluate the second integral on the RHS, we write
x x x
2
2
2
1
1
2
3
2
+ + = +
+
( ).Step 4
Replace
x +
1
by y
1
1
3
2
2 2
2
2
2
( )
.
x x
dx
dx
y
+ +
∫
=
+
∫
Using the reduction formula described above, we get
1
1
1
2
2
3
4
1
1
2
3
4
1
2
2 2
2
( )
( )
x x
dx
x
x x
dx
x
+ +
∫
=
+
⋅ + +
+
⋅
+
+
∫
=
+
+ +
+
+
2
2
2
2
3
2
2 1
3 1
2
3
1
2
x
x x
dx
x
( )
++
∫
=
+
+ +
+
+
−
3
2
2 1
3 1
2
3
2
3
1
2
3
2
2
2
1
x
x x
x
( )
tan
=
+
+ +
+
+
−
2 1
3 1
4
3 3
2 1
3
2
1
x
x x
x
( )
tan
∴
−
+ +
∫
=
−
+ +
−
+
+ +
+
+
−
2 3
1
1
1
4
2 1
3 1
4
3 3
2 1
3
2 2 2 2
1
x
x x
dx
x x
x
x x
x
( ) ( ) ( )
tan
=
−
+ +
+
− +
+ +
−
( )
( )
ta ...