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Applied maths/AL/tutorial/De-0/sol/p.1
PLK Vicwood K.T.Chong Sixth Form College
Applied Mathematics (AL)
Solution to Tutorial
Topic: Methods of integration
x
dx
1. (a)
x 1
1
(1 
)dx
=
x 1
= x  ln |1 + x| + C
Code: De - 0


(b)
6 x 2  11x  5
dx
3x  1
2
(2 x  3 
)dx
=
3x  1
2
= x2 3x + ln |3x  1| + C
3
where C is an arbitrary constant.


where C is an arbitrary constant.
(2 x  3) 2 2
 ln |3x  1| + C0
4
3
9
where C0 = C  is an arbitrary constant.)
4
(or

(c)
=
=
(or
=
2 x 2  10 x
dx
x 2 4
10 x
8
(2 

)dx
2
2
x 4 x  2 2
x
2x + 5 ln|x2 + 4|  4tan1( ) + C
where C is an arbitrary constant.
2
x
x
2x + 10 ln |sec (tan1( ))|  4tan1( ) + C0
2
2
x
x
2x + 10 ln 1  ( ) 2  4tan1( ) + C0
2
2

x
x2
= 2x + 5 ln ( 1 
)  4tan1( ) + C0
2
4
where C0 = C + 5ln4 is an arbitrary constant.)
(d)

2x 2
dx
x 2 9
18
=
(2 
)dx
2 2
x 3
x
= 2x  6tan1( ) + C
3

where C is an arbitrary constant.
Applied maths/AL/tutorial/De-0/sol/p.2
2.
(a) Putting u = ex,
du
dx 1
 .
= ex = u; i.e.,
dx
du u
 1  e x
dx
=
 (1  1 )u
du
u

du
=
u 1
= ln |u +1| + C
= ln |ex +1| + C
where C is an arbitrary constant.
(b) Putting u = x3 + 1,
x
=
1
3
5
dx
1
du

= 3x2; i.e.,
.
dx
du 3x 2
( x 3  1) 4 dx
 (u  1)u
4
du
u6 u5
+C

18 15
( x 3  1) 6 ( x 3  1) 5
=
+C
where C is an arbitrary constant.

18
15
du
2x
dx x 2  1
2

(c) Putting u = x + 1,
; i.e.,
.

dx x 2  1
du
2x
xdx
=
 ( x 2  1) ln( x 2  1)
=
 2u
du
1
ln |u| + C
2
1
=
ln |ln (x2+1)| + C
where C is an arbitrary constant.
2
1
dx
du 1  56
(d) Putting u = x 6 ,
= 6u5.
 x ; i.e.,
du
dx 6
dx
=
x
=
1
2
1
 x3
6u 5 du
 u3  u 2

1
)du
u 1
= 2u3  3u2 + 6u  ln |u + 1| + C
=
6 (u 2  u  1 
1
1
1
1
= 2 x 2  3 x 3 + 6 x 6  ln | x 6 + 1| + C
where C is an arbitrary constant.
Applied maths/AL/tutorial/De-0/sol/p.3
x
3. (a)

A
B

x  1 ( x  1) 2
for x 1.
x 2  2x  1
x = A(x  1) + B = A x A + B
By comparing coefficients,
A=1
and
A + B = 0.
Therefore, A = B = 1.
x
dx
(b)
2
x  2x  1
1
1
(

)dx
=
x  1 ( x  1) 2
for all real x.


= ln |x  1| 
1
+C
x 1
where C is an arbitrary constant.
4. (a) Note that
x3
 x 3
2
7x  6
.
( x  1)( x  2)
x  3x  2
Therefore, A = 1, B = 3.
7x  6
C
D


for all real numbers x  1, 2.
2
x  3x  2 x  1 x  2
C(x + 2) + D(x + 1) = 7x + 6
for all real numbers x.
Putting x = 1, we have C = 1.
Putting x = 2, we have D = 8; i.e., D = 8.
x3
 x 2  3x  2 dx
(b)
=
=

(x  3 
1
8

)dx
x 1 x  2
1 2
x  3x  ln |x| + 8ln |x + 2| + C
2
xe x
dx
(1  x ) 2
1
xe x d (
)
1 x
where C is an arbitrary constant.

5. (a)
=

=
xe x


1 x
=

=
=
xe x

1 x
 ( 1  x )d ( xe
1

x
)
e x dx
xe x
 ex + C
1 x
ex
+C
1 x

where C is an arbitrary constant.
Applied maths/AL/tutorial/De-0/sol/p.4
 sec
(b)
=

3
xdx
sec xd tan x
= sec x tan x 

tan xd sec x
= sec x tan x 

sec x tan 2 xdx
= sec x tan x 
 sec x(sec
= sec x tan x 


=
=
2
x  1)dx
sec 3 xdx 

sec xdx .
sec xdx
1
sec x tan x +
2
1
sec x tan x +
2

1
sec xdx
2
1
ln |sec x + tan x| + C
2
where C is an arbitrary constant.