TTBDKTPT VQ

GV: PHAN VĂN VINH

‹ CÁC BÀI TỐN TÍCH PHÂN LUYỆN THI ĐẠI HỌC‹
1/ Cho hàm số : f(x)= x.sinx+x2 . Tìm nguyên hàm của hàm số g(x)= x.cosx
biết rằng nguyên hàm này triệt tiêu khi x=k π
2/Định m để hàm số: F(x) = mx 3 +(3m+2)x2 -4x+3 là một nguyên hàm của hàm số:
f(x) = 3x2 +10x-4.
3/Tìm họ nguyên hàm của hàm số: f(x)= cos 3 x.sin8x.
3.TÍNH :
π
3

2

4/I = ∫ 3tg x dx

∫ sin x dx

12 / I =

π
2 3

13/I =

∫

2

5/I = ∫ (2cotg x + 5) dx
π
6
π
2 1 − cos x

π
3

0

∫ 1 + cos x dx

3

7/ I = ∫ sin2 x.cos2xdx

1
∫π 2 x 2 x dx
sin cos
4
2
2

15/I =

0

π


π

3

4

∫

2

2

(2cos x-3sin x)dx

16/I =

0

− π
2

π
− x )
4
dx
π
s in (
+ x)
4


cotg2x dx

17/I = ∫ esin

−

2

x

sin 2x dx

π

3

∫π

e tgx+ 2
2
I= ∫
cos
x
0
4

(tgx-cotgx)2 dx

6


18/

π
2

π
4

11/ I = ∫ cos 4 x dx
0

2

∫

4
4
21/I = cos2x(sin x + cos x)dx

π
2

22/ I = ∫ cos3 xdx
0
π
2

4sin 3 x
23/ I = ∫
dx

0 1 + cosx
1

3
2
∫ x 1 − x dx

0
1

25/I = ∫ x 5 1 + x 2 dx

x
dx
2x
+
1
0
1
1
27/I = ∫ x
dx
0e +4
2
1
dx
28/I = ∫
−x
11− e
2

e2x
29/I = ∫ x
dx
e
+
1
0
1
e− x
dx
30/I = ∫ − x
e
+
1
0
e
ln x
dx
31/I = ∫
2
x(ln
x
+
1)
1
26/I = ∫

π
4


π

dx

0
1

π
2

s in (

10 / I =

∫
π

x

π

24/ I =

6

9/ I=

∫

0


π

2

6

0

4
∫ sin x dx

14/I =

π

π
2

sin 3 x − sin x
cot gx dx
sin x

1

∫ cos

20/ I =

π

2

0

8/I =

4

3

0

π
4
π
4

6/I =

π

π
2

1
19/ I = ∫ sin 4 x dx
π
4

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ


DĐ: 0905.838.969

1


TTBDKTPT VQ

7
3

0
2

3

0
1

34/I =

1

∫
3

x

2


4−x
1

4

35/I = ∫

2

36*/I =

x 16 − x
6
1

∫

2 3

2

2

dx

dx

2

x x −9


dx

−1
2

38/I = ∫ x (x 2 + 4)3 dx
0

x2 − 4
dx
x

∫

39/I =

4 3
3

− 2

∫

40*/I =

−2
ln 2

41/I =


∫

x2 +1

dx
2
x x +1
e x − 1dx

0
1

1
42/I = ∫
dx
3
−
2x
0

ln x 3 2 + ln 2 x
dx
48/I = ∫
x
1
e
sin(ln x)
dx
49/I = ∫

x
1
e

1

3

4

5

5

43/I = ∫ sin xdx
0
π
3

1
44*/I = ∫
dx
cos
x
0
e−2x
dx
45/I = ∫ − x
e
+

1
0
ln 3
1
dx
46/I = ∫
0
ex + 1

dx
ln 2 e − 1
e 2
x +1
62/I = ∫
.ln xdx
x
1

61/I =

1

51/I = ∫ (1 + 2x)(1 + 3x + 3x 2 )3 dx

1

52/I = ∫

1x
π

3

1+ x

3

π
2

64/I = ∫ sin x.sin 2x.sin 3xdx
0
π
2

65/I = ∫ cos 2x(sin 4 x + cos 4 x)dx
0

π
2

dx

66*/I = ∫ ( 3 cos x − 3 sin x )dx
0

2

2

53/I = ∫ tg x + cot g x − 2dx

π
6
1

x7
67/I = ∫
dx
8
4
1
+
x
−
2x
2
3

π
2

68*/I = ∫ 4cos x − 3sin x + 1 dx

54/I = ∫ (1 − x 2 )3 dx

0

0

1
55*/I = ∫ 2x

dx
e
+
3
0
ln 3
ex

∫

x

(e + 1)

0

3

69/I = ∫ x. 3 1 − xdx
1
2

dx

0

57/I = ∫ x(e2x + 3 x + 1)dx
π
2


58/I = ∫ 6 1 − cos3 x sin x.cos5 xdx
0

2 3

59*/I =

∫

5

1
x x2 + 4

π
4

x
dx
1
+
cos
2x
0

60/I = ∫

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

4sin x + 3cos x + 5


9

1

56/I =

x

x2
dx
63/I = ∫
(x
+
1)
x
+
1
0

0

2

∫

1

0


−1

π
2

1

dx

50/I = ∫ x (x − 1) dx

2
2
∫ x 4 − x dx

4

sin 2 x cot gx

π
6

2

37/I =

1

47/I = ∫


33/I = ∫ (x − 3) x 2 − 6x + 8 dx

e2x

ln 5

π
4

x +1
dx
3
3x + 1

32/I = ∫

GV: PHAN VĂN VINH

x +1
dx
3
0 3x + 2
π
x
71*/I = ∫ sin 6 dx
2
0
2
x
72*/I = ∫

dx
2
+
x
+
2
−
x
0

70/I = ∫

3

73/I =

dx

3
2
∫ x . 1 + x dx
0
1

ln(1 + x)
dx
2
x
+
1

0

74**/I = ∫
π
2

sin x
dx
sin
x
+
cos
x
0

75/I = ∫

DĐ: 0905.838.969

2


TTBDKTPT VQ

2

eπ

76/I = ∫ cos(ln x)dx
1

2

77*/I = ∫ 4 + x 2 dx
0
2

x
dx
1
+
x
−
1
1

78/I = ∫
e

79/I = ∫
1
3

GV: PHAN VĂN VINH

1 + 3ln x ln x
dx
x

90*/I = ∫ ln( 1 + x − x)dx
0

3

x −1
x +1
dx
92/I = ∫
x
1
3
x3
93/I = ∫ 2
dx
x
−
16
1

82/I = ∫
e
e2

83/I = ∫
1
2

cos x
dx
2
6
−

5sin
x
+
sin
x
0

94/I = ∫

85/I =

ln x
dx
x
ln x
dx
ln x

86/I = ∫
0

87/I =

1

∫ x 2 + 3 dx
3

1


1
4 − x2

dx

π2
4

∫ sin xdx

0
π
3 ln(sin x)

88/I = ∫

π
6
2

cos 2 x

e2

dx

89/I = ∫ cos(ln x)dx

96/I =


1
1
−
)dx
ln 2 x ln x

∫

x 2ex
dx
110*/I = ∫
2
(x
+
2)
0
1

π

111/I = ∫ e2x sin 2 xdx
0
2

cos 2x + 1dx

99/I = ∫ cos x

1
e


ln x
dx
2
(x
+
1)
1

sin xdx

113/I = ∫

0
2π

e
1
2

∫ 1 + sin xdx

0
3π
4

114/I = ∫ x.ln
0

∫ sin 2x dx


t

102/I = ∫ 1 − sin xdx

2

π
3

0
3

103/I = ∫  ln(x + x 2 + 1)  dx


−1

1

1+ x
dx
1− x

 ln x 
115/I = ∫ 
 dx ⇒ I < 2
x



1

π
4
π

1

1
x

112/I = ∫ x 2 ln(1 + )dx

π
4
π

101/I =

∫ x cos xdx

0

97/I = ∫ x 3 − 2x 2 − x + 2 dx
−1
3π
4

108/I =


π2
4

109/I = ∫ x.sin x cos 2 xdx

2
∫ x − 4 dx

100/I =

∫ x sin xdx

0
π
6

−4
2

98/I =

π2
4
0

e

84/I = ∫ x ln(x 2 + 1)dx
1
3


107/I =

3

1
e2

dx

104*/I = ∫

π
6

95*/I = ∫ (

81/I = ∫ (ln x) 2 dx

2

2
8 3

80/I = ∫ ln(x 2 − x)dx
2
e

1


∫

91*/I =

x sin x
dx
2
0 1 + cos x
1
1
105*/I = ∫ 2
dx
x
(x
+
1)(4
+
1)
−1
1
x4
106*/I = ∫
dx
x
1
+
2
−1
π


2

116/I = ∫ sin x.ln(cos x)dx
0
π
e2

117/I =

2
∫ cos (ln x)dx

1
ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

DĐ: 0905.838.969

3


TTBDKTPT VQ

π
4

1
dx
cos
x
0


118/I = ∫

GV: PHAN VĂN VINH

1
dx
3
cos
x
0

1

2

120/I = ∫ x 3e x dx
0
π
2

2

121/I = ∫ esin x .sin x cos3 xdx
0
π
2

sin 2x
122/I = ∫

dx
4
0 1 + cos x
1
3
123/I = ∫ 2
dx
x
−
4x
−
5
0
2
5
dx
124/I = ∫ 2
x
−
6x
+
9
1
1
1
dx
125/I = ∫
2
2x
+

8x
+
26
−5
1
2x + 9
126/I = ∫
dx
x
+
3
0
4
1
dx
127/I = ∫ 2
x
(x
+
1)
1
0
sin 2x
128*/I = ∫
dx
2
−π (2 + sin x)
2
1


x −3
129/I = ∫
dx
2
0 (x + 1)(x + 3x + 2)
1

4x
dx
3
(x
+
1)
0
1
1
131/I = ∫ 4
dx
2
(x
+
4x
+
3)
0

130/I = ∫

π
3


3

sin x
dx
2
(sin
x
+
3)
0

132/I = ∫

4sin x
dx
π 1 − cos x

145/I = ∫ x 1 − xdx

6
π
3

146/I = ∫

3

133/I = ∫


π
4

119*/I = ∫

1

π
3

0
6

4

1
134/I = ∫
dx
2
cos
x.sin
x
π

0

−1
3

6

π
3

148/I = ∫

135/I = ∫ sin x.tgxdx

1
2

0
π
3

−1
2

139/I =

π
3

140/I =

152/I =
1
dx
2
2
π sin x + 9cos x


∫

153/I =

cos x − 1

1 + sin x

cos x
dx
0 sin x + cos x + 1
4
1
dx
142/I = ∫ 2
x
(x
+
1)
1
141/I = ∫

−3

1
x + 4 + (x + 4)3

π
3


sin 3 x
dx
144/I = ∫
0 cos x

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

∫

1 + e2x
1

2
7 x 9+x

dx

dx

0
π
2

0
π
2

1


∫

0
4

3e4x + e2x

154/I = ∫ e x sin 2 xdx

∫ 1 + 3cos x dx

143/I = ∫

1
2

dx

π
2

∫ cos x + 2 dx
π
−
2
π
2

2


x + 4x + 13
1
151/I = ∫
dx
x
0 3+ e
−2
1

sin3 x
dx
137/I = ∫ 2
2
5
(tg
x
+
1)
.cos
x
0

3
π
2

2x − 5

∫


150/I =

4
π
4

−

4x − x 2 + 5 dx

149/I = ∫

1
dx
136/I = ∫
sin
2x
π

138/I =

∫

147/I =

x−4 1
.
dx
x+2 x+2
1

dx
x 2 + 2x + 9
1
dx
4x − x 2

dx

cos 4 x
155/I = ∫
dx
4
4
cos
x
+
sin
x
0
1
3
dx
156/I = ∫
0 x+9 − x
π

157/I = ∫ x sin xdx
0
π


158/I = ∫ x 2 cos 2 xdx
0
1

159/I = ∫ cos x dx
0

DĐ: 0905.838.969

4


TTBDKTPT VQ

1

160/I = ∫ sin x dx

GV: PHAN VĂN VINH

2

0

161/I =

1
2

π2

4

∫ x sin x dx
0

2

162/I =

π
4

ln x
176/I = ∫ 5 dx
1 x
e
ln x
dx
177/I = ∫
2
(x
+
1)
1

∫ x cos x dx

0
π


2

163/I = ∫ x cos x sin x dx

164/I =

0
π
6

2

∫ x cos x sin x dx

0
4

x

165/I = ∫ e

e
1
2

0
π
2
π
3


180/ ∫ esin

0
π

181/I=
2x

0
1

2

sin x dx
2 x

x e
168/I = ∫
dx
2
(x
+
2)
0
e

169/I = ∫ (1 + x) ln x dx
1
e


2

170/I = ∫ x ln x dx
1
1
e

2

171/I = ∫ ln x dx
1
e

172/I = ∫ x(2 − ln x) dx
1

173/I = ∫ (
e
2

189/I = ∫

1
1
−
)dx
2
ln x ln x


174/I = ∫ (x 2 + x) ln x dx

x

e +e

0
e

190/I=

dx

∫ ln x dx
π
2

191/I = ∫ (esin x + cos x) cos x dx

2

x

sin x cos3 x dx

π
2

sin 2x.cos x
dx

0 1 + cos x

192/I = ∫

π
2

sin 2x + sin x
dx
1 + 3cos x
0

193/I = ∫

π
4 1 − 2sin 2

x
dx
1 + sin 2x

194/I = ∫

0
π
2

−x

1

e

0

166/I = ∫ e3x sin 4x dx

e2

1+ x
dx
1− x

179/I = ∫ cos x.ln(1 − cos x) dx
π
2

dx

ex

1

0

178/I = ∫ x ln

1
π
4


167/I = ∫ e

1
x

175/I = ∫ x 2 ln(1 + ) dx

sin 2x

∫ 1 + sin 4 x dx

0
π
2

sin 2x
dx
182/I = ∫
4
0 1 + cos x
2
5
183/I = ∫ 2
dx
x
−
6x
+
9
1

1 2
x + 3x + 2
dx
184/I = ∫
x
+
3
0
4
1
185/I = ∫ 2
dx
x
(x
+
1)
1
1
ln(1 + x)
dx
186/I = ∫ 2
x
+
1
0
1
1+ x4
187/I ∫
dx
6

1
+
x
0
1

188/I = ∫ x15 1 + x 8 dx

3

∫

195/I =

2

x +1

0
π
3

196/I = ∫

π
4

x 5 + 2x 3

dx


tgx
2

cos x 1 + cos x

2

197/I = ∫ (
−1
π
4

dx

x −1 2
) dx
x+2

198/I = ∫ x.tg 2 x dx
0
5

199/I = ∫ ( x + 2 − x − 2 ) dx
−3
4

200/I = ∫

−1

2

201/I = ∫
1

2
dx
x +5 +4
x
dx
x+2 + 2−x

0

1
ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

DĐ: 0905.838.969

5


TTBDKTPT VQ

GV: PHAN VĂN VINH

2

ln(1 + x)
202/I = ∫

dx
x2
1
π
2

sin 2x
203/I = ∫
dx
0 1 + cos x
π
2

sin 2008 x
dx
2008
x + cos 2008 x
0 sin

204/I = ∫

216/I =

2
2

x

∫


π
2

2

dx

2

1− x
1− x2
dx
217/I = ∫
4
11+ x
0
2

7

218/I =

∫

x
3

1
2


π
4

0
ln 2

3

sin x
dx
2
0 cos x

207/I = ∫

π
2

2

208/I = ∫ cos x.cos 4x dx
0
1

1
dx
2x
x
e
+

e
0
e
ln x
210/I = ∫
dx
2
1 (x + 1)

209/I = ∫

e
1

1
211/I = ∫
dx
x
+
1
+
x
0
1
x2
dx
212/I = ∫
2
4
−

x
0
1
x
213/I = ∫
dx
2
0 4−x
1
2

x4
dx
214/I = ∫ 2
0 x −1
π
2

sin 3x
215/I = ∫
dx
0 cos x + 1

1

2
0 x − 3x + 2

232*/I = ∫ x sin x.cos 2 xdx


2

dx

0
π
2

cos x
dx
0 cos 2x + 7
4
1
234/I = ∫ 2
dx
x
(x
+
1)
1
233/I = ∫

π
2

221/I = ∫ x + 1dx

2
3
235/I = ∫ sin 2x(1 + sin x) dx


0
π
2

222/I = ∫ (cos3 x + sin 3 x) dx
0
3

x2 +1
223/I = ∫
dx
+
x
1
0

∫

2

x x +9

dx

π

238/I = ∫ x sin 3 x cos 4 xdx
0


cos x
cos 2 x + 1

dx

x +1
226/I = ∫ 3
dx
3x
+
1
0
π
2 1 + sin 2x

+ cos 2x
dx
cos x + sin x

227/I = ∫

π
6

x 2

1

(1 + e )
dx

2x
1
e
+
0

228/I = ∫

3

x +1
dx
3
3x
+
2
0
4
1

236/I = ∫

7

224/I = ∫ (1 + x) 2 .e2x dx

0
7
3


0
2

237/I =

1

225/I = ∫

dx

π

3

2

0
π
2

4x − 1

231/I = ∫

1+ x
1 − ex
205/I = ∫ sin x.ln(1 + cos x) dx
219/I = ∫
dx

x
0
+
1
e
0
2
3
1
x +1
206/I = ∫
dx
220/I = ∫ x 1 − x dx
2
x
1
0
π
2

sin x.cos3 x
dx
230/I = ∫
2
cos
x
+
1
0


2

3

229/I = ∫ x (1 − x) dx
0

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

π
2

239/I = ∫ cosx cosx − cos3 xdx
−

π
2

1

240*/I = ∫ ln( x 2 + a + x)dx
−1
π
2

1 − sin x
dx
x
(1
cos

x)e
+
0

241/I = ∫

π
2

sin 2x + sin x
dx
+
cos3x
1
0

242/I = ∫

π
4

sin 2x
dx
2
2
sin
x
+
2cos
x

0

243/I = ∫

DĐ: 0905.838.969

6


TTBDKTPT VQ

244/I =

2
2

∫

0

245/I =

2
2

∫

0
1


246/I =

∫

2
2
1

247/I = ∫
0

2

248/I =

∫

2
3
1

x

π
3

3

1− x2
x3

1− x

GV: PHAN VĂN VINH

2

dx

256/I = ∫ tg 4 xdx

dx

π
2 1 + sin x

257*/I = ∫

0 1 + cos x

1− x
dx
x2
2

4 − x2
1
2

dx


x x −1

dx

0
π
2

sin x
dx
1
+
sin
x
0

250/I = ∫

π
2

cos x
dx
7
+
cos
2x
0
4
1

252/I = ∫
dx
2
(1
+
x)x
1
2
x +1
dx
253/I = ∫ 3
0 3x + 2
251/I = ∫

π
3

1

e x dx

2 3

258/I = ∫ (1 − x ) dx
0
π
4

259/I = ∫ x.tg 2 xdx


cos x + sin x
dx
3 + sin 2x
π

1
dx
2 2
(4
+
x
)
0
1
3x 2
261/I = ∫
dx
3
0 x +2
2
1 − x5
262*/I = ∫
dx
5
1 x(1 + x )
260/I= ∫

π
3


cos x
dx
2
0 1 − sin x

263/I = ∫

π
3

sin 2 x
264/I = ∫
dx
6
0 cos x
π
6

sin x + sin 3 x
265/I = ∫
dx
cos 2x
0
π
2

1
dx
π sin x 1 + cos x


254*/I = ∫

266/I = ∫

4

3
π
2

255/I = ∫ cosx cosx − cos3 xdx
−

0
π
4

sin 4 x − cos 4 x
270/I = ∫
dx
sin
x
+
cos
x
+
1
0
π
4


sin 4 x − cos 4 x
271/I = ∫
dx
sin
x
+
cos
x
+
1
0
π
2

sin x cos x + cos x
dx
sin x + 2
0

272/I = ∫

0
2

249/I = ∫ x 5 (1 − x 3 )6 dx

π
2


269/I = ∫ sinxcosx(1+cosx)2dx

π
4

2

x

π
2

sin x
dx
2
cos
x
+
3
0

267/I = ∫

π
2

π2

268/I =


∫
0

sin x
dx
x

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

1

273/I = ∫

1
ex
3

dx

x
x 3 + 2x 2 + 10x + 1
274/I = ∫
dx
2
x + 2x + 9
0
1
x3
dx
275/I = ∫ 2

3
0 (x + 1)
1
3
276/I = ∫ 3
dx
x
+
1
0
1 4
x +1
dx
277*/I = ∫ 6
+
x
1
0
1
x
278/I = ∫
dx
3
(2x
+
1)
0
7
1
dx

279/I = ∫
2 2 + x +1
a
1

280/I =

3
2

∫

1
2
1

281*/I = ∫
0

1
x 1− x

2

dx

x ln(x + 1 + x 2 )
1+ x

2


dx

4

282/I = ∫ (x − 1)2 ln x dx
1
DĐ: 0905.838.969

7


TTBDKTPT VQ

3

GV: PHAN VĂN VINH

283/I = ∫ x ln(x + 1)dx

296/I =

0
2

3x 3
284/I = ∫ 2
dx
1 x + 2x + 1
1

4x − 1
285/I = ∫ 3
dx
2
+
+
+
x
2x
x
2
0
1
2

1

286/I = ∫

−1 (3+ 2x)
2

1

2

5+12x +4x

1


287/I = ∫

x + 1+ x

0
π
2

dx

dx

cos x
dx
2 + cos 2x

288/I = ∫

0
π
2

cos x + sin x
dx
3
+
sin
2x
π


289/I = ∫

4
π
2

290/I = ∫ (cos3 x + sin 3 x)dx
0
π
2

291/I = ∫ cos5 x sin 4 xdx
0
π
2

292/I = ∫ cos2x(sin4 x+cos4 x)dx
0
π
2

1
dx
2
+
sin
x
0

293/I = ∫


π
2

1
dx
2
−
cos
x
0

294/I = ∫

2

295/I =

∫

2
3

1
2

x x −1

dx


∫

3

0
2

297*/I = ∫
1
1

298/I = ∫

1+ x2
1

3

π
2

sin x
dx
0 cos x + sin x

dx

+ 1+ x2
1


−1 1 +
π
3

1
dx
2x
−1 3 + e

309*/I = ∫

dx

x 1+ x
x3

0x
1

299/I = ∫

1

x3

7

2

310*/I = ∫

π
2

dx

x + 1+ x2

dx

1
300/I = ∫
dx
4
π sin x cos x
6
π
2

cos x
dx
cos
x
+
1
0

π
2

cos x

dx
302/I = ∫
2
−
cos
x
0
π
2

sin x
303/I = ∫
dx
sin
x
+
2
0
π
2

cos3 x
304/I = ∫
dx
cos
x
+
1
0


306/I =

π
2

tgx

312*/I = ∫

1 − ln 2 (cos x)

0

dx

π
2

sin x
dx
cos
x
+
sin
x
0
1
1
dx
314*/I = ∫ x

2
−1 (e + 1)(x + 1)
313*/I = ∫

301/I = ∫

305/I =

sin 4 x
311/I = ∫
dx
4
4
0 cos x + sin x

π
2

1

∫ 2cos x + sin x + 3 dx
0
π
2

cos x
dx
∫
2
(1

−
cos
x)
π
3

π
4

307/I = ∫ tg 3 x dx
0

sin 2 x
dx
308*/I= ∫ x
3
+
1
−π
π

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

1

315*/I = ∫ e
0
1

3x +1


dx

x2

316*/I = ∫
0
π
2

2

x +4

dx

cos3 x
dx
317*/I = ∫ 4
2
0 cos − 3cos x + 3
318*/Tìm x> 0 để
π
3

319*/I = ∫

π
4
1


t 2et
dt = 1
∫
2
(t
+
2)
0

x

tan x
2

cos x cos x + 1

dx

320*/I = ∫ −3x 2 + 6x + 1dx
0
π
4

321*/I = ∫ tg 5 x dx
0

DĐ: 0905.838.969

8



TTBDKTPT VQ
1

π
4

336,Cmr

322/I = ∫ cotg 3 x dx
π
6
π
3

50

4

1
2

dt

∫ ( 3+ 2cost )

2

0


339, Cmr

1

∫ 2 + tgx dx

∫
1
2

2x −1
x −1
dx < ∫
dx
x
x +1
1

2 3

341, I =

∫

5

π
3


x x2 + 4

cos 2x
dx
2
π 1 − cos 2x

342, I = ∫ x3 1 − x2 dx

6

sin 2x + sin x
dx
343, I = ∫
1 + 3cos x
0

0

π
2

π
4

327*/I = ∫ ( t g x − 1 ) 2 d x
tg x + 1

1


328*/I =

1
2

2 3

1

(e x + 1) e x − 1

π
−1
e4

1
e

3

332/I = ∫
1

dx
346, I = ∫ 2
x + x +1
0

1
dx

x cos 2 (ln x + 1)

1
ln 5

348, I =

1 + 3ln x ln x
dx
x

dx
∫ln3 ex + 2e− x − 3
sin 2x cos x
dx
1 + cos x
0
3

350, I = ∫ ln ( x 2 − x ) dx

π
4

2
π
2

333*/I = ∫ ln(1 + tgx)dx
π

2

π
dx
π
≤∫
≤
2
16 0 5 + 3cos x 10

0
1

(

−7

)

3

358, I =

dx

∫ x+x
1
π
4


359, I =

1 − cos 2xdx

∫
π
−
3
2

360, I = ∫

3

5 ( x − 1)

dx

1

x2 − x − 6

1

x ln x + 1 + x 2

361, I = ∫

(


1 + x2

0

e2

ln x + ln ( ln x )
x

π
2

)dx

dx

3

363, J = ∫ cos2 x dx
π
6

sin x

1

x2 dx
x6 − 9
0


364, I = ∫
π
3

dx

365, I = ∫

4

π
4

sin x cos 5 x
3

π
2

sin 2x
dx
sin x + 6 sin 2 x + 5
0

366, I = ∫

4

π


367 I = ∫ sin xex dx
π
2

368, I = ∫ cos 5 x co s 7x d x
0

352, I = ∫ ( x − 2 ) e dx
2x

0

2

54 2 ≤ ∫ x+7+ 11−x dx ≤108

1 − 2sin 2 x
dx
+
1
s
in2x
0

357, I = ∫

0

351, I = ∫ ( esin x + cos x ) cos xdx


0

0

π
4

π
2

349, I = ∫

1
dx
x 6 (1 + x 2 )

11

e

347, I = ∫

356, I = ∫ x sin xdx

e

1

dx


0

π2

362, I = ∫

dx
345, I = ∫
1 + x2
0

ex

ln 3

331/I = ∫

0
1

x − x3
dx
x4

329*/I = ∫

0

344, I = ∫ 1 − x 2 dx


x
dx
3
x +1

∫

330/I = ∫

1

2

355, I = ∫ x3ex dx

dx

1

326/I = ∫

)

2

2

340, I = ∫

5

325/I = ∫ sin x dx
0 cos x + 1

0

(

2 3+ 3

x
dx
1 1+ x −1

0

1

1

≤

2
2
≤ ∫ e x − x dx ≤ 2 4 e
2
e
0

2


π
2

335,Cmr

354, I = ∫

2

338,Cmr

π
4

324*/I =

x
dx
1 + cos 2x
0

1 ≤ ∫ 2 dx ≤ 4

337,Cmr 1 ≤

π
4

π
4


x3

−1

323/I = ∫ tg x dx

334,Cmr

GV: PHAN VĂN VINH

353, I = ∫ x − x dx
2

0

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

369, I =

π
4

sin 4 x + cos4 x
∫π 3x + 1 dx

−

4


DĐ: 0905.838.969

9


TTBDKTPT VQ

GV: PHAN VĂN VINH

AN ACT OF KINDNESS IS NEVER WASTED !

ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ

DĐ: 0905.838.969

10