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Chapter 5e Integral of Irrational Function
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1
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8/
( )
2
dx
I
x q ax bx c
=
− + +
∫
( )
( )
( )
(
)
2
2
2
2 2
2
2 2 2
2 2
dx 1 dt 1
Put x q dx , t
t x q
t
x q ax bx c
1 1 1 2q 1
ax bx c a q b q c a q b q c
t t t t
t
a b 2aq 1
aq bq c a t b 2aq t aq bq c
t
t t
− = ⇒ = − =
−
− + +
+ + = + + + + = + + + + +
÷
÷ ÷ ÷
+
= + + + + = + + + + +
∫
( )
( )
( )
(
)
( )
( )
(
)
( )
( )
(
)
2
2
2 2
2
2
2 2 2 2
2
dt
dx
t
1 1
x q ax bx c
a t b 2aq t aq bq c
t
t
dt
dt
t
1
a t b 2aq t aq bq c t aq bq c t b 2aq a
t
−
⇒ =
− + +
+ + + + +
−
= = −
+ + + + + + + + + +
∫ ∫
∫ ∫
2
( )
( )
( ) ( )
( )
( ) ( )
( ) ( ) ( )
( )
2
2
2 2 2
2
2
2
2 2
2 2
2
dt
Put aq bq c A,
b 2aq
1 a
t t
aq bq c aq bq c aq bq c
dt
b 2aq b 2aq
1 a
t
A 2A A
4A
b 2aq b 2q 4Aa b 2aq
a
Put t y dy dt, N
2A A
4A
4 aq bq c
= − + + =
+
÷
+ +
÷
+ + + + + +
÷
= −
+ +
÷
+ + −
÷
÷
+ + − +
+ = ⇒ = = − =
÷
+ +
∫
∫
( ) ( )
( )
( )
( )
( )
(
)
2
2
2 2
2 2
2 2 2
2
2
2 2
2
2
dy Ady
I If N 0 N n 4Aa b 2aq ,
y N y N
A
4Aa b 2aq 4a a.q b.q c b 4abq 4 aq 4ac b
b c b
A 0 A m a.q b.q c 0 a q 0
2a a 4
b 4ac b
a q a 0
2a 4
= − = − ≥ ⇒ = ⇒ ≥ +
+ +
≥ + ⇔ + + ≥ + + ⇔ ≥
> ⇒ = ⇒ + + ≥ ⇔ + + − ≥
÷
−
⇔ + + ⇔ >
÷
∫ ∫
( )
( )
( )
2 2
2 2
2
m.dy
I m.ln y y n
y n
B 1 b 2q
y t
2A x q
2 aq bq c
⇒ = − = − + +
÷
+
+
= + = +
÷
−
+ +
∫
3
( ) ( ) ( )
( ) ( )
( )
( )
(
)
( )
( )
( )
( )
(
)
2
2
2 2 2
2
2 2
2 2
2
2 2
2
2
2
b 2aq b 2aq b 2aq
a a
y n t t t
2A A
4A
aq bq c aq bq c
t aq bq c t b 2aq a
aq bq c
t aq bq c t b 2aq a
t
t
aq bq c
+ + +
÷
+ = + + − = + +
÷
÷
+ + + +
÷
+ + + + +
=
+ +
+ + + + +
÷
=
÷
+ +
÷
( )
( )
(
)
( )
( )
2 2
2
2
2
2
2
2 2 2
2
2
t aq bq c t b 2aq a
*
t
1 2q 1 1 1
a q b q c a q b q c
t t t t
t
ax bx c
ax bx c y n
x q aq bq c
1 1
x q t
t x q
+ + + + +
= + + + + + = + + + +
÷
÷ ÷ ÷
+ +
= + + ⇒ + =
− + +
= + ⇒ =
÷
−
( )
( )
( )
( )
( )
2 2
2
2
2
2
2 2
ln y y n
dx
I
1
x q ax bx c
aq bq c
1 b 2aq 1 ax bx c
ln aq bq c
x q x q
2 aq bq c aq bq c
+ +
÷
⇒ = = −
− + +
+ +
+ + +
÷
÷
= − + + + +
÷
÷
− −
+ + + +
÷
÷
∫
4
( )
2
dx
* I
x 1 1 x
=
− −
∫
( )
2
2
2
2 2
dx 1
* I Put : x 1
t
x 1 1 x
dt 1 1 2t
dx 1 x 1 1
t
t t
= − =
− −
− +
⇒ = − = − + = −
÷
∫
( )
( ) ( )
2
2
1
1
2
dt
dt
t
I
1 2t 1 2t
t. t
because1 x 0 x 1 x 1 0 t 0 t t
d 2t 1 1 2t
1 1 2 1 x
I 1 2t 1 C
1
2 2 x 1 1 x
1 2t
1
2
− +
⇒ = − =
− − − −
− > ⇔ < ⇒ − < ⇒ < ⇔ = −
− − − −
−
= − = − = − − − = − − − = − +
− +
− −
− +
∫ ∫
∫
( )
( )
( )
( )
(
)
( )
2
2
2
2 2
2
2 2 2
2 2
2
3x 4
VD : dx x 6x 8 1 x 3
x 6x 8
Put x 3 t x t 3 dx dt
3 3 t 4
3x 4
dx dx
x 6x 8 1 t
d 1 t
3t.dt dt 3
13 13arcsin t
2
1 t 1 t 1 t
d 1 t 2t.dt 3 1 t 13arcsin t
3 x 6x 8 13arcsin x 3 C
+
− + − = − −
− + −
− = ⇒ = + ⇒ =
+ +
+
=
− + − −
−
= + = − +
− − −
− = − = − − +
= − − + − + − +
∫
∫ ∫
∫ ∫ ∫
For evaluating integral
2
R x, ax bx c
+ +
÷
∫
we can make a trigonometric change of
variables:
5
( )
( )
( )
2
2
2
2
2
2
2 2 2
2
2
2
2 2 2
2
2
2
2 2 2
2
b c b
ax bx c a x
2a a
4a
b b 4ac
a x a u d If b 4ac 0, a 0
2a
4a
b 4ac b
a x a u d If b 4ac 0, a 0
2a
4a
b 4ac b
a x a d u If b 4ac 0, a 0
2a
4a
÷
+ + = + + −
÷
÷
−
÷
= + − = − − > >
÷
÷
−
÷
= + + = + − < >
÷
÷
−
÷
= − − + = − − − > <
÷
÷
2 2 2
2 2 2
2 2 2
d
R x, ax bx c R u, u d Put u
sin t
R x, ax bx c R u, d u Put u d.sin t
R x, ax bx c R u, u d Put u d.tgt
+ + = − =
÷ ÷
+ + = − =
÷ ÷
+ + = + =
÷ ÷
∫ ∫
∫ ∫
∫ ∫
*
( )
( )
3
2
x 2 3
dx 1
I sin arctan C
3 3
x 4x 7
+
= = +
÷
÷
+ +
∫
( )
( )
( )
( ) ( )
( )
2
2
3 3
2 2
2 2
2
3
2 3
dx du
* I x 4x 7 x 2 3 Put u x 2 I
x 4x 7 u 3
3 3
Put u 3.tan t du , u 3 3 tan t 1
cos t
cos t
x 2 3
3.cos t.dt sin t 1 u 3 1
I sin arctan sin arctan
3 3 3 3 3
cos t. 3
= + + = + + = + ⇒ =
+ + +
= ⇒ = + = + =
+
⇒ = = = =
÷
÷
÷
∫ ∫
∫
6
( )
( )
( )
2
2
3 2
2
2
b
x
dx 4a
2a
* I sin arctan with a 0, 4ac b 0
a. 4ac b
4ac b
ax bx c
a
÷
+
÷
÷
= = > − ≥
÷
−
−
÷
+ +
÷
∫
( )
( )
( )
( ) ( )
2
2
2 2
2
3
2
2
2
2
3
2 2
2 2 2 2
2
dx b 4ac b
* I a 0, 4ac b 0 ax bx a a x
2a
4a
ax bx c
b 4ac b 1 du
Put u x , m I ,
2a
a. a
4a
u m
m.dt m
Put u m.tan t du , u m m tan t 1
cos t
cos t
−
= > − ≥ + + = + +
÷
÷
÷
+ +
−
= + = ⇒ =
÷
÷
+
= ⇒ = + = + =
∫
∫
( )
( )
3
3 2 2 2
3
2 2
2
2 2
2
2
2
1 du 1 m.dt 1 cos t.dt 1 sin t
I .
a. a a. a a. a
a. a.m cos t. m
m
u m
cos t.
cos t
b
x
1 u 4a
2a
sin arctan sin arctan
m
4ac b a. 4ac b
4ac b
a. a.
4a
a
⇒ = = = =
+
÷
÷
+
÷
÷
= =
÷
÷
− −
−
÷
÷
÷
÷
∫ ∫ ∫
( )
( )
( ) ( )
c
1
2
3
c
2b
b
2a
2a
2 2
c
b
x
dx 4a
2a
I lim .sin arctan
m
a. 4ac b
ax bx c
b
c
4a 4a
2a
. lim sin arctan
m
a. 4ac b a. 4ac b
+∞
→+∞
−
−
→+∞
+
÷
÷
÷
⇒ = =
÷
−
+ +
÷
+
÷
÷
÷
= =
÷
− −
÷
∫
7
( )
( )
( )
( ) ( )
( )
2
2
2 2
2
n
2
2
2
2
n n
2 2
2 2 2 2
2
n n n
2 2
2
dx b 4ac b
* I a 0, 4ac b 0 ax bx a a x
2a
a
ax bx c
b 4ac b 1 du
Put u x , m I ,
2a
a
a
u m
m.dt m
Put u m.tan t du , u m m tan t 1
cos t
cos t
1 du 1 m.dt
I
m
a a
u m
cos t.
cos t
−
= > − ≥ + + = + +
÷
÷
÷
+ +
−
= + = ⇒ =
÷
÷
+
= ⇒ = + = + =
⇒ = =
+
∫
∫
∫
( )
n 2
n n 1
n
cos t .dt
1
m
a
−
−
=
÷
∫ ∫
( )
( )
( )
n 2
2
n 1
n n
2
2
2
2
cos t .dt
dx 1
* I a 0, 4ac b 0
m
a
ax bx c
4ac b b u
m , u x , t arctan
2a m
a
−
−
= = > − ≥
+ +
−
= = + =
÷
÷
∫ ∫
( )
2
2 2
2 2
2 2
b
x
dx 1 dt t 1
2a
n 2 : I arctan
a m a.m
ax bx c
4ac b 4ac b
a.
4a 4a
2 2ax b
arctan
4ac b 4ac b
÷
÷
+
÷
÷
= = = = =
÷
÷
+ +
÷
− −
÷
÷
÷
+
÷
÷
=
÷
÷
− −
∫ ∫
( )
( )
2
2 2 3 2 3
2
2 3 2 3
dx 1 1 1 cos 2t
n 4 : I cos t .dt .dt
2
a .m a .m
ax bx c
1 sin 2t 1 u 1 u
t arctan .sin 2.arctan
2 m 2 m
2a .m 2a .m
b
x u 0, x u
2a
+
= = = =
+ +
= + = +
÷ ÷ ÷
÷
= − ⇒ = = +∞ ⇒ = +∞
∫ ∫ ∫
8
( )
( )
( )
( )
2 2 3
2
0
b/2a
3/2 3/2
2 2
2
2
2
3
3/2
2
dx 1 u 1 u
. arctan .sin 2.arctan
m 2 m
2a .m
ax bx c
1 sin 1 sin
. .
2 2 2 2
4ac b 4ac b
2a .
2a .
4a
2a
4a sin
.
2 2
4ac b
+∞
+∞
−
⇒ = +
÷ ÷
÷
+ +
π π π π
= + = +
÷ ÷
− −
÷
÷
π π
= +
÷
−
∫
( )
(
)
3
n
3/2
2
n
2
2
i 1
c
c.n b
* lim 1 sin arctan
2 4ac b
a i.c b.n.i.c n c
→+∞
=
→+∞
÷
÷
= −
÷
÷
−
+ +
∑
( )
( )
[ ]
( )
(
)
( )
i i i
3
2
i i
3
n n
2
3/2
3
n n
2
2
2
i 1 i 1
2
c c
2
n
3/2
n n
2
i 1
c c
1 c i.c
Put f x , x 0, c , x , x
n n
ax bx c
c.n c 1
I lim lim
n
a i.c b.n.i.c n c
a i.c b.n.i.c n c
n
c 1
lim lim
n
i.c i.c
a b c
n n
→+∞ →+∞
= =
→+∞ →+∞
→+∞ →+∞
=
→+∞ →+
= ∈ ∆ = =
+ +
⇒ = =
+ +
+ +
÷
÷
= =
÷
+ +
÷ ÷
÷
∑ ∑
∑
( )
( )
c
n
i
3/2
c
2
i 1
0
2
dx
f x . x lim
ax bx c
b
1 sin arctan
2 4ac b
→+∞
=
∞
∆ =
+ +
÷
÷
= −
÷
÷
−
∑
∫
( )
p
m n
x a bx dx+
∫
with m, n, p is rational number. The Russian mathematicant Trebushep
prove that the upper integral only can be expressed in elementary function in 3 follow cases:
1/ p is an interger, when that, put
s
x t=
with s is the least common multiple of m, n.
2/
m 1
n
+
is an interger, put
s s
a bx t+ =
with s is the denominator of p.
3/
m 1
p
n
+
+
is an interger, put
n s
ax b t
−
+ =
with s is the denominator of p.
9
( )
( )
( )
( )
( )
( )
( )
( )
( ) ( )
( ) ( )
10
1 1
2 4
10
4
4 3
3
10 10 9 10
2
8 9 8 9
4 4
dx
Ex : I x x 1
x x 1
So p 10 is a interger, we have case 1/
Put x t dx 4t .dt
t 1 1 d t 1 d t 1
4t .dt
I 4 dt 4
t t 1 t 1 t 1 t 1
4 4 1 4
8 t 1 9 t 1
2 x 1 9 x 1
−
−
÷
= = +
÷
÷
+
= −
= ⇒ =
+ − + +
÷
⇒ = = = −
÷
+ + + +
− −
= + = +
+ +
+ +
∫ ∫
∫ ∫ ∫ ∫
( )
( )
( )
( )
( )
( )
p
m n
p
a a 1 m.a n.a a 1
p p
i
a m 1 1 n.a.i a m 1 1
i n.a p i i i p i
p p
i 1 i 1
* I x . a b.x .dx where p is a interger
Put x t dx a.t .dt I a t . a b.t .t .dt
I a t . C b.t .a .dt a C t .b a .dt
Put n.a.i a m 1 1 c
− −
+ − + + −
− −
= =
= +
= ⇒ = ⇒ = +
= =
÷ ÷
÷ ÷
+ + − = ⇒
∫
∫
∑ ∑
∫ ∫
i c 1 i p i
p p
p
i c i p i
p
i 1 i 1
C t .b a
I a C t .b a .dt a.
c 1
+ −
−
= =
= =
÷
÷
+
∑ ∑
∫
( )
( ) ( )
( )
( ) ( )
i n.i m 1 i p i i n.i m 1 i p i
p p
p
p p
m n
i 1 i 1
1
i n.i m 1 i p i i i p i
1
p p
p
p p
m n
1
i 1 i 1
0
0
C x .b a C x .b a
I x . a b.x .dx a.
a n.i m 1 n.i m 1
C x .b a C .b a
I x . a b.x .dx
n.i m 1 n.i m 1
+ + − + + −
= =
+ + − −
= =
= + = =
+ + + +
⇒ = + = =
+ + + +
∑ ∑
∫
∑ ∑
∫
10
( )
( )
( )
( )
c
i n.i m 1 i p i
c
p
p
p
m n
1
c c
i 1
1
1
c
c
f
i i p i i i p i
p p
p p
f
c c
i 1 i 1
1
1
i i
p
If n, m 0, n.i m 1 0 n.i m 1 f
C x .b .a
I lim x . a b.x .dx lim
n.i m 1
C x .b .a C .b .a
lim lim
n.i m 1
x n.i m 1
C .b .a
+ + −
→+∞ →+∞
=
−
− −
→+∞ →+∞
= =
< + + < ⇒ + + = −
= + =
+ +
= =
+ +
+ +
= −
∑
∫
∑ ∑
( )
( )
p i i i p i
p p
p
i 1 i 1
C .b .a
n.i m 1 n.i m 1
because n.i m 1 0
− −
= =
=
+ + + +
+ + <
∑ ∑
( )
( )
( )
p
1
m n
n
p
m n
n
i 1
0
i i p i
p
p
i 1
1 i i
* lim . a b. x . a b.x .dx
n n n
C .b a
m, n, p is interger 0
n.i m 1
→+∞
=
−
=
÷
+ = +
÷ ÷
÷
= >
+ +
∑
∫
∑
( )
( )
( )
[ ]
( )
( )
( )
p
m n
i i i i i
p
1
m n
n n
p
m n
i
n n
i 1 i 1
0
i i p i
p
p
i 1
1 i
* Put f x x . a b.x m, n, p is interger 0 , x 0, 1 , x , x
n n
1 i i
I lim . a b. lim f x . x x . a b.x .dx
n n n
C .b a
n.i m 1
→+∞ →+∞
= =
−
=
= + > ∈ ∆ = =
÷
⇒ = + = ∆ = +
÷ ÷
÷
=
+ +
∑ ∑
∫
∑
11
( )
(
)
( )
( )
i i p i
p
n
p
p
n
m m 1
1
n
1
i 1 i 1
h
i i p i
p
p
1
1
i 1
C .b a
* lim i h . a b. i.h
n .i m 1
C .b a
with n .i m 1 0
n .i m 1
−
+
→+∞
= =
→+∞
−
=
+ = −
+ +
= + + <
+ +
∑ ∑
∑
( )
( )
( )
[ ]
( )
( )
(
)
( ) ( )
p
n
m
1
i i i 1 1 i
n
p
n
m m 1
1
i
n
i 1
h
p
m 1 n
1
n
m
n
i 1
c
n
c
* Put f x x . a b.x m, n , p is intergers, p 0, m,n 0 , x 1, c ,
i c 1
c 1
x , x I lim i h . a b. i.h
n n
c 1 i c 1
c 1
lim i . a b. Dat h
n n n
c 1
lim
n
+
→+∞
=
→+∞
+
→+∞
=
→+∞
→+∞
→+∞
= + > < ∈
−
−
∆ = = ⇒ = + =
− −
−
÷
= + =
÷ ÷
÷
−
=
÷
∑
∑
( ) ( )
( )
( )
( )
p
m n
1
n
i 1
i i p i
c
p
n
p
p
m n
i
n c
1
i 1 i 1
1
c
i c 1 i c 1
. a b.
n n
C .b a
lim f x . x lim x . a b.x .dx
n .i m 1
=
−
→+∞ →+∞
= =
→+∞
− −
÷
+
÷ ÷
÷
= ∆ = + = −
+ +
∑
∑ ∑
∫
We have the formula:
( )
( ) ( )
n
2
n 1
2 2
P x dx
dx
Q x ax bx c p. 1
ax bx c ax bx c
−
= + + +
+ + + +
∫ ∫
( )
n
P x
is a n degree polynomial,
( )
n 1
Q x
−
is a n – 1 degree polynomial with indefinite
coefficients
To determine p and coefficients of
( )
n 1
Q x
−
, we take derivative (1) and equate coefficients of
two sides to obtain system of equations.
(How to prove this formula?)
12
*
( )
3
2 2 2
2
x x 1 1 5
I dx 2x 5x 1 x 2x 2 .ln x 1 x 2x 2
6 2
x 2x 2
− +
= = − + + + + + + + +
÷
+ +
∫
( )
( )
( )
( )
( ) ( )
( )
3
2 2
2 2
3
2 2
2 2 2
3 2 2
3 2 2
x x 1 dx
VD : I dx ax bx c x 2x 2 d.
x 2x 2 x 2x 2
Take derivative in two side, we obtain :
x x 1 2x 2 d
2ax b x 2x 2 ax bx c
x 2x 2 2 x 2x 2 x 2x 2
x x 1 2ax b x 2x 2 ax bx c x 1 d
2ax 4ax 4ax bx 2bx 2b a
− +
= = + + + + +
+ + + +
− + +
= + + + + + + +
+ + + + + +
⇔ − + = + + + + + + + +
= + + + + + +
∫ ∫
( ) ( ) ( )
3 2 2
3 2
,
x bx cx ax bx c d
3ax x 5a 2b x 4a 3b c 2b c d
3a 1, 5a 2b 0 4a 3b c 1, 2b c d 1
+ + + + + +
= + + + + + + + +
⇒ = + = + + = − + + =
( )
( )
( )
( )
3
2
2 2
2
2
2 2
2
2
3
2 2 2
2
1 5 1 5 x x 1
a , b , c , d I dx
3 6 6 2
x 2x 2
1 5 dx
2x 5x 1 x 2x 2 .
6 2
x 2x 2
d x 1
dx
ln x 1 x 2x 2
x 2x 2
x 1 1
dx
ln x x a
x a
x x 1 1 5
I dx 2x 5x 1 x 2x 2 .ln x 1 x 2x 2
6 2
x 2x 2
− +
⇒ = = − = = ⇒ =
+ +
= − + + + +
+ +
+
= = + + + +
÷
+ +
+ +
= + +
÷
+
− +
⇒ = = − + + + + + + + +
+ +
∫
∫
∫ ∫
∫
∫
÷
13
*
( )
1
n
n
n n
0
dx 1
I
2
1 x 1 x
= =
+ +
∫
( )
( )
( )
1 n
1 1
n
n
n
n n
0 0
n
n n n
1 n n
n
n
dx
I 1 x dx
1 x 1 x
1 1 z
Put: x 1 z x 1 x 1
z 1 z 1
z 1
+
−
−
= = +
+ +
+ = ⇔ = ⇔ + = + =
− −
−
∫ ∫
( )
n 1 n 1
n 1 n 1 n 1 n 1
n 1 n 1
n
n
n
1
n
n
x 0
z dz z dz
n.x dx n.z dz dx x .z dz
x
z 1
x 1 z 2
change the limits
1
x 0 z lim 1
x
− −
− − − + −
− − +
+
→
⇒ − = ⇔ = − = − = −
−
= ⇒ =
= ⇒ = + = +∞
÷
( )
( )
( )
( ) ( )
1
1 1
1
n
n
n
n n
0 0
1 n
n
2
n n 1
n
n n 1
a
a
n
n
n n
n
2 2
1 n n 1
2
2
1 n n 1
a a a
a a
a
n n
n n
n n
dx
So I 1 x dx
1 x 1 x
z z dz
lim .
z 1
z 1
z .z dz 1
lim lim z dz lim
z
z 1 . z 1
1 1 1
a
2 2
− −
+
−
−
+
→+∞
− − −
−
+ +
→+∞ →+∞ →+∞
−
= = +
+ +
= −
÷
÷
−
−
= − = − =
− −
= − =
∫ ∫
∫
∫ ∫
14
*
dx
I
x 1 x 1 2
=
+ + − +
∫
( )
2
2
3
dx
15/ I
x 1 x 1 2
x 1 x 1 2
Put u x 1 x 1 0
x 1 x 1 x 1 x 1
2 2 4
x 1 x 1 2 x 1 u 4 x 1 u 4
u u
u
8
4dx 2u du
u
=
+ + − +
+ − +
= + + − = = >
+ − − + − −
⇔ + − − = ⇒ + = + ⇔ + = + +
⇒ = −
÷
∫
( )
( )
( )
4 4
3 3 2
4 3 4 3 2
2
1
2 2
2 2
1
2
2
2 2
2
2 2 2
8 4 4
2u u
1 1 u 4 1 u
u u u
I du du du du du
4 u 2 2 u 2 2 2
u 2u u 2u u 2u
4
2 2.dk 2
u
I du put u du , k
k u
u 2u k
k 2.dk 1 k
I dk
4 4
2 1 k
k
k
k
2u 2
u u 2u 2u 2 2
I 1
u 2u u 2u u 2u
− −
−
= = = = −
+ +
+ + +
= − = ⇒ = − =
+
⇒ = − − =
÷
+
+
− +
+ − + −
= = = +
+ + +
∫ ∫ ∫ ∫ ∫
∫
∫ ∫
2
2
u 2u
+
+
( )
( )
( )
2 2
2
2
2 2
2
2
1 u 1 k
I du dk
2 2 1 k
u 2u
d 1 k
1 2u 2 2 1 k 1
du du du dk
2 4 1 k 1 k
u 2u u 2u
1 1 1
u ln u 2u 2. du k 1 dk ln 1 k
2
2 4
1 .u
u
⇒ = +
+
+
+
+ −
= − + + +
+ +
+ +
= − + + + − + +
+
÷
∫ ∫
∫ ∫ ∫ ∫ ∫
∫ ∫
15
3
2
2
2
2
2
d 1
2 1 2du 2
u
I du ln 1
2 2 2
u
u
1 .u 1 1
u u u
1 2 1 k
I u ln u 2u ln 1 k ln 1 k C
2 u 4 2
2
Where 0 u x 1 x 1
k
+
÷
−
= = − = − = − +
+ + +
÷ ÷ ÷
⇒ = − + − + + − + + +
< = = + + −
∫ ∫ ∫
( )
( )
( ) ( )
( )
( )
( ) ( )
( )
( )
2
2 2 2
2
2 2 2
2
2 2 2
1
1 1 1
2
2
2
2
1
d x 1
2x 1 dx 1 3
* I dx J J
10
x 1
x 1 x 1 x 1
x 1 t
dx
J Put x tan t dx 1 tan t dt 4
x 2 t arctan 2
x 1
+
+
= = + = − + = +
+
+ + +
π
= ⇒ =
= = ⇒ = + ⇒
= ⇒ =
+
∫ ∫ ∫
∫
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
2
arctan 2 arctan 2
2
2
2
/4 /4
arctan 2
arctan 2
/4
/4
1 tan t dt
J cos t dt
tan t 1
1 cos 2t sin 2t
t
d 2t
4 2 4
sin 2.arctan 2
arctan 2
1
2 4 8 4
sin 2.arctan 2
arctan 2
1
So I
20 2 4 8
π π
π
π
+
⇒ = =
+
+
= = +
π
= + − −
π
= + + −
∫ ∫
∫
16
4
1
dx
* I
x x
=
+
∫
( )
( )
4
1
4 2 2 2
2 2 2
1 1 1 1
2
2 2
2
2
1
2 2
1
1 1
2 2 2
2
2
2
2
1
1 1 1
x 1 u 1
dx dx
* I put x u du
x 4 u 2
x x 2 x
x 2u 2u 1 1
I 2. dx du du du
x x .2 x
u u u u u u
d u u
2u 1
I du ln u u ,
u u u u
du
1
d 1
1 1
u
u
I du Ln 1
1 1
u
u u
1 1
u u
I
= ⇒ =
= = ⇒ = ⇒
= ⇒ =
+
+
⇒ = = = −
+
+ + +
+
+
= = = +
+ +
−
+
÷
= − = = = +
+
+ +
⇒
∫
∫ ∫ ∫ ∫
∫ ∫
∫ ∫ ∫
2
2
2
1
1
1 3 3 9
Ln u u Ln 1 du Ln6 ln 2 Ln ln 2 ln 3 ln Ln
u 2 4 4
= + + + = − + − = + =
( ) ( )
( )
( ) ( )
( )
( ) ( )
( ) ( )
2
2
4 2
2 2
2 2
2 2 2 2
2 2
3 3
2 x 1 x x dx
2 2
x 1 1
I
2
x x 1
x 1 3.x x 1 3.x
x 1 3x dx x 1 3x dx
1 1
2 2
x 1 3.x x 1 3.x x 1 3.x x 1 3.x
1 dx 1 dx
2 2
x 1 3.x x 1 3.x
+ − +
÷
+
= =
− +
+ − + +
+ − + +
= +
+ − + + + − + +
= +
+ + + −
∫ ∫
∫ ∫
∫ ∫
17
( ) ( )
2 2
2 2
1 dx 1 dx
2 2
3 1 3 1
x x
2 2 2 2
3 3
x x
1
2 2
.2 arctg arctg C
1 1
2
2 2
arctg 2x 3 arctg 2x 3 C
= +
+ + − +
÷ ÷
÷ ÷
+ −
÷
÷ ÷
÷
÷ ÷
= + +
÷
÷ ÷
÷ ÷
÷
= + + − +
∫ ∫
*
x
2
2 2
0
arcsin x x. 1 x
I 1 u du 1 x dx C
2 2
−
= − = − = + +
∫ ∫
We have: I = The area of plane figure bounded by the graph
2
y 1 x= −
and Ox.
And
( )
2 2 2
y 1 x y x 1 choose y 0= − ⇔ + = >
is a half circle with radius R = 1.
We have the follow figure:
We have: x cos sin arcsin x
2
π
= − α = α ⇒ α =
÷
I = area of yellow part = area of sector AOB + area of triangle BOC
18
2 2
x
2
2 2
0
x. 1 x arcsin x x. 1 x
I
2 2 2 2
arcsin x x. 1 x
I 1 u du 1 x dx C
2 2
α − −
= + = +
−
⇒ = − = − = + +
∫ ∫
*
( )
2 2
x a x a x a
I dx x a a.ln
x a
x a x a
− − + +
= = − − +
+
− − +
∫
( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )
( )
2 2 2
' '
2 2 2 2 2
2
2
2
2 2
2 2 2
2 2 2
2
2 2 2
2
2 2 2
x a x a
* I dx put t t a x x a x t 1 a 1 t
x a x a
a 1 t a 1 t t 1 a 1 t t 1
x dx dt
t 1
t 1
2at t 1 2at 1 t
4a.t.dt 2t.dt
dt I a 2t.
t 1 t 1 t 1
d t 1
2t.dt 2t.dt 1
put u 2t, dv v
t
t 1 t 1 t 1
− −
= = ⇒ + = − ⇔ − = − +
+ +
− + − + − + + −
⇔ = ⇒ =
−
−
− − + +
= = ⇒ =
− − −
−
= = ⇒ = = = −
−
− − −
∫
∫
∫ ∫
( )
, du 2dt,
1
=
( ) ( ) ( )
( )
2 2 2
2 2
2at dt 2at 1 t 1
I 2a 2a. .ln
2 t 1
t 1 t 1 t 1
x a x a x a
2a 1 2a
x a x a x a
x a x a x a
I dx a.ln a.ln
x a 2a
x a
x a x a x a
1
1
x a x a
x a
x a x a x a x a x a x a
a . a.ln x a a.ln
x a a
x a x a x a x
+
⇒ = − + = − +
−
− − −
− − −
+
− − + +
+ + +
⇒ = = − + = +
− −
+
− − − +
−
−
÷ ÷
+ +
+
− + − + + − + +
= − + = − − +
+
− − + − − +
∫
∫
a
The second method:
19
( )
( )
(
)
( )
( )
(
)
2 2
2 2
2
2
2
x a 1 x
I dx put x a.cos 2t dx 2a.sin2t.dt t .arccos
x a 2 a
x a a cos 2t 1 a 1 2sin t 1 2a.sin t
x a a 1 cos 2t a 1 2cos t 1 2a.cos t
2a.sin t i.sin t
I 2a sin2t.dt 2a .2sin t.cos t.dt
cos t
2a.cos t
4a.i sin t.dt 4a
−
= = ⇒ = − ⇒ =
+
− = − = − − = −
+ = + = + − =
−
⇒ = − = −
= − = −
∫
∫ ∫
∫
1 cos 2t
i dt 2ai dt 2ai cos 2t.dt 2ai.t ai.sin2t
2
x ai 2x
ai.arccos .sin arccos
a 2 a
−
= − + = − +
= − +
÷
∫ ∫ ∫
2 2
a x 1 2x x
I dx a.i .sin arccos arccos
a x 2 a a
x a x a
x a a.ln
x a x a
−
⇒ = = −
÷
÷
+
− + +
= − − +
− − +
∫
( ) ( )
a x a x
* I dx a x a x 2a.arctan
x a a x
− −
= = − + −
÷
+ +
∫
( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )
( )
( ) ( )
( ) ( ) ( )
( ) ( )
( )
( )
( )
2 2 2
' '
2 2 2 2 2
2
2
2
2 2
2 2 2
2 2 2
2
2 2 2
2
2 2 2
a x a x
* I dx put t t a x a x x t 1 a 1 t
x a x a
a 1 t a 1 t 1 t a 1 t 1 t
x dx dt
1 t
1 t
2at 1 t 2at 1 t
4a.t.dt 2t.dt
dt I a 2t.
1 t 1 t 1 t
d 1 t
2t.dt 2t.dt 1
put u 2t, dv v
1 t
1 t 1 t 1 t
− −
= = ⇒ + = − ⇔ + = −
+ +
− − + − − +
⇔ = ⇒ =
+
+
− + − −
−
= = ⇒ = −
+ + +
+
= = ⇒ = = = −
+
+ + +
∫
∫
∫ ∫
, du 2dt=
20
( )
( ) ( ) ( )
2 2 2
2t 2dt 2at
I a a 2a.arctgt
1 t 1 t 1 t
a x a x
2a 2a
a x a x a x
a x a x
I dx 2a.arctg 2a.arctg
a x 2a
a x a x a x
1
a x a x
a x a x a x
a x 2a.arctg a x a x 2a.arctg
a x a x a x
− −
⇒ = − − − = −
÷
+ + +
− −
− − −
+ +
⇒ = = − = −
÷ ÷
−
+ + +
+
+ +
− − −
= + − = − + −
÷ ÷
+ + +
∫
∫
( )
( )
(
)
( )
( )
(
)
2 2
2 2
a x
The second method: I dx
a x
1 x
put x a.cos 2t dx 2a.sin2t.dt t .arccos
2 a
a x a 1 cos 2t a 1 1 2sin t 2a.sin t
a x a 1 cos 2t a 1 2cos t 1 2a.cos t
−
=
+
= ⇒ = − ⇒ =
− = − = − − =
+ = + = + − =
∫
2
2
2
2a.sin t sin t
I 2a sin2t.dt 2a .2sin t.cos t.dt
cos t
2a.cos t
1 cos 2t
4a sin t.dt 4a dt 2a dt 2a cos 2t.dt
2
x a 2x
2at a.sin2t a.arccos .sin arccos
a 2 a
a x x a 2x
I dx a.arccos .sin arccos a
a x a 2 a
⇒ = − = −
−
= − = − = − +
= − + = − +
÷
−
⇒ = = − + =
÷
+
∫ ∫
∫ ∫ ∫ ∫
∫
2 2
a x
x 2a.arctg
a x
−
− −
÷
+
21
( )
2001 2
2000
2004
x .dx x 1 x 1 x 1
* I
x 1 2003 x 1 1001 x 1 2001
x 1
÷
= = − +
÷ ÷ ÷
÷
+ + +
+
∫
( )
( )
( )
2000 2
2000
2004 2
2000 2
2003 2
2
2000 2002 2001 2000
x .dx t 1 1 1
* I put t x 1 dx dt I .dt
t t
t
x 1
t 1 1 1 1 1 1 x
d dat u 1 du d u 1
t t t t t x 1 x 1
u 2.u
I u u 1 .du u 2u u .du
2003
−
= = + ⇒ = ⇒ = −
÷
÷ ÷
+
−
= − − = − ⇒ = − ⇒ = − =
÷ ÷ ÷ ÷
+ +
⇒ = − = − + = −
∫ ∫
∫
∫ ∫
( )
002 2001
2001 2
2000
2004
u
2002 2001
x .dx x 1 x 1 x 1
I
x 1 2003 x 1 1001 x 1 2001
x 1
+
÷
⇒ = = − +
÷ ÷ ÷
÷
+ + +
+
∫
( )
a 1 2
a
a 4
x .dx x 1 x 2 x 1
* I
x 1 a 3 x 1 a 2 x 1 a 1
x 1
+
+
÷
= = − +
÷ ÷ ÷
÷
+ + + + + +
+
∫
( )
( )
( )
a 2
a
a 4 2
a 2
a 3 a 2 a 1
2
a a 2 a 1 a
a
x .dx t 1 1 1
* I put t x 1 dx dt I .dt
t t
t
x 1
t 1 1 1 1 1 1 x
d dat u 1 du d u 1
t t t t t x 1 x 1
u 2.u u
I u u 1 .du u 2u u .du
a 3 a 2 a 1
x .d
I
+
+ + +
+ +
−
= = + ⇒ = ⇒ = −
÷
÷ ÷
+
−
= − − = − ⇒ = − ⇒ = − =
÷ ÷ ÷ ÷
+ +
⇒ = − = − + = − +
+ + +
⇒ =
∫ ∫
∫
∫ ∫
( )
a 1 2
a 4
x x 1 x 2 x 1
x 1 a 3 x 1 a 2 x 1 a 1
x 1
+
+
÷
= − +
÷ ÷ ÷
÷
+ + + + + +
+
∫
22
2
dx
* I
ax bx c
=
+ +
∫
( )
2 2
2
2
2
2 2
2
2 2
dx 1 dx
* I
a
ax bx c
b c b
x
2a a
4a
1 dx
if b 4ac 0
a
b b 4ac b b 4ac
x x
2a 2a
1 4a .dx
a
2a.x b b 4ac 2a.x b b 4ac
= =
+ +
+ + −
÷
= − >
+ − − −
÷ ÷
+ +
÷ ÷
=
+ + − + − −
÷ ÷
∫ ∫
∫
∫
( ) ( )
2 2
2 2
2 2
2 2
2 2
dx
*
2a.x b b 4ac 2a.x b b 4ac
A B
2a.x b b 4ac 2a.x b b 4ac
1 A 2a.x b b 4ac B 2a.x b b 4ac
A B 0 A B, b A B b 4ac B A 1 2B b 4ac 1
1 1
B , A
2 b 4ac 2 b 4ac
+ + − + − −
÷ ÷
= +
+ + − + − −
÷ ÷
⇒ = + − − + + + −
÷ ÷
⇒ + = ⇒ = − + + − − = ⇒ − =
⇒ = = −
− −
2 2
2 2 2 2
2 2
2 2 2
4a.dx
I
2a.x b b 4ac 2a.x b b 4ac
4a.dx 4a.dx
2 b 4ac 2a.x b b 4ac 2 b 4ac 2a.x b b 4ac
2a.d 2a.x b b 4ac 2a.d 2a.x b b 4ac
2a 2a
b 4ac 2a.x b b 4ac b 4ac 2
⇒ =
+ + − + − −
÷ ÷
= − +
− + + − − + − −
÷ ÷
+ + − + − −
÷ ÷
= − +
− + + − −
÷
∫
∫ ∫
∫
2
a.x b b 4ac
+ − −
÷
∫
23
2 2
2
2
2
2 2
ln 2a.x b b 4ac ln 2a.x b b 4ac
b 4ac
dx 1 2a.x b b 4ac
I ln
ax bx c
b 4ac 2a.x b b 4ac
− + + − + + − −
÷ ÷
=
−
+ − −
⇒ = =
+ +
− + + −
∫
( )
(
)
( )
(
)
2
2 2
2 2
2
2
2
2
2
2
2
2
2a 1 1
Check the solution :
b 4ac
2a.x b b 4ac 2a.x b b 4ac
2a.x b b 4ac 2a.x b b 4ac
2a
2a.x b b 4ac
b 4ac
2a 2 b 4ac 4a
4a
2a.x b b 4ac
b 4ac
÷
− +
÷
÷
−
+ + − + − −
÷ ÷
÷
− + − − + + + −
÷ ÷
÷
=
÷
÷
+ − +
−
÷
−
÷
= =
÷
+ − +
−
÷
( )
2
2
1
ax bx c
.x 4ab.x 4ac
=
+ +
+ +
( )
2 2
2
2
2
2
2
2
2 2 2 2
dx 1 dx
* I
a
ax bx c
b c b
x
2a a
4a
1 dx
if b 4ac 0
a
b 4ac b
x
2a 2a
b
x
1 2a 2 2ax b
2a
. .arctan .arctan
a
4ac b 4ac b 4ac b 4ac b
2a
= =
+ +
+ + −
÷
= − <
−
÷
+ +
÷
÷
÷
+
+
÷
÷
= =
÷
÷
− − − −
÷
÷
∫ ∫
∫
24
( )
2
2 2 2
2
2
b
d x
dx 1 dx 1
2a
* I if b 4ac 0
a a
ax bx c
b c b b
x x
2a a 2a
4a
1 2
I
b
2ax b
a x
2a
+
÷
= = = − =
+ +
+ + − +
÷ ÷
−
⇒ = − =
+
+
÷
∫ ∫ ∫
( )
2
2
dx
* I
ax bx c
=
+ +
∫
( )
( )
( )
2
2
2
2 2
2
2
2
1
2a.x b b 4ac
2a
dx
* I dx
1
b 4ac
ax bx c
2a.x b b 4ac
if b 4ac 0
−
÷
+ + −
÷
÷
÷
= =
÷
−
+
+ +
÷
÷
+ − −
÷
÷
− >
∫ ∫
( )
(
)
2 2
2 2
2
2 2
2
2
2
d 2a.x b b 4ac d 2a.x b b 4ac
4a
2a 2a.x b b 4ac 2a 2a.x b b 4ac
b 4ac
2dx
2a.x b b 4ac
+ + − + − −
÷ ÷
÷
+
÷
÷
+ + − + − −
÷ ÷
÷
=
÷
−
÷
−
÷
+ − +
÷
∫ ∫
∫
25
