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𝑑𝑢
1
1
𝟏
1. ∫ 𝑒 5−4𝑥 𝑑𝑥 = |𝑢 = 5 − 4𝑥, 𝑑𝑢 = −4𝑑𝑥| = ∫ 𝑒 𝑢 −4 = − 4 ∫ 𝑒 𝑢 𝑑𝑢 = − 4 𝑒 𝑢 + 𝐶 = − 𝟒 𝒆𝟓−𝟒𝒙 + 𝐶.
2. ∫ sin(𝜋𝑧) 𝑑𝑧 = |𝑢 = 𝜋𝑧, 𝑑𝑢 = 𝜋𝑑𝑧| = ∫ sin 𝑢
3. ∫ cos 2 (3𝜃) 𝑑𝜃 = ∫
1+cos(6𝜃)
2
𝟏
𝑑𝑢
𝜋
1
1
𝟏
= 𝜋 ∫ sin 𝑢 𝑑𝑢 = − 𝜋 cos 𝑢 + 𝐶 = − 𝝅 𝐜𝐨𝐬(𝝅𝒛) + 𝐶.
𝟏
𝑑𝜃 = 𝟐 𝜽 + 𝟏𝟐 𝐬𝐢𝐧(𝟔𝜽) + 𝐶.
1
1
1
𝟏
𝟏
4. ∫ 𝑥𝑒 −2𝑥 𝑑𝑥 = |𝑢 = 𝑥, 𝑑𝑢 = 𝑑𝑥, 𝑑𝑣 = 𝑒 −2𝑥 𝑑𝑥, 𝑣 = − 2 𝑒 −2𝑥 | = − 2 𝑥𝑒 −2𝑥 + 2 ∫ 𝑒 −2𝑥 𝑑𝑥 = − 𝟐 𝒙𝒆−𝟐𝒙 − 𝟒 𝒆−𝟐𝒙 + 𝐶.
5. ∫ 𝑡 3 ln 𝑡 𝑑𝑡 = |𝑢 = ln 𝑡 , 𝑑𝑢 =
𝑑𝑡
𝑡
1
4𝑥
6. ∫ √9−4𝑥 2 𝑑𝑥 = |𝑢 = 9 − 4𝑥 2 , 𝑑𝑢 = −8𝑥𝑑𝑥| = − ∫
𝑑𝑥
2
2
1
1
, 𝑑𝑣 = 𝑡 3 𝑑𝑡, 𝑣 = 4 𝑡 4 | = 4 𝑡 4 ln 𝑡 − 4 ∫ 𝑡 4
𝑡
𝟏
𝟏
= 𝟒 𝒕𝟒 𝐥𝐧 𝒕 − 𝟏𝟔 𝒕𝟒 + 𝐶.
= −√𝑢 + 𝐶 = −√𝟗 − 𝟒𝒙𝟐 + 𝐶.
√𝑢
3
𝑑𝑢
2
3
7. ∫ √9−4𝑥 2 = |𝑢 = 3 𝑥, 𝑑𝑢 = 3 𝑑𝑥, 𝑥 = 2 𝑢| = ∫
2𝑑𝑢
𝑑𝑡
1
2
√9−4(3𝑢)
𝑑𝑢
1
𝟏
𝟐
= 2 ∫ √1−𝑢2 = 2 arcsin 𝑢 + 𝐶 = 𝟐 𝐚𝐫𝐜𝐬𝐢𝐧 (𝟑 𝒙) + 𝐶.
2
𝑒 𝑡 −𝑒 −𝑡
sinh 𝑡
8. ∫ 𝑒 𝑡 +𝑒 −𝑡 𝑑𝑡 = ∫ cosh 𝑡 𝑑𝑡 = |𝑢 = cosh 𝑡 , 𝑑𝑢 = sinh 𝑡 𝑑𝑡| = ∫
cos 𝛼
𝑑𝑢
𝑢
= ln|𝑢| + 𝐶 = ln|cosh 𝑡| + 𝐶 = ln(cosh 𝑡) + 𝐶 = 𝐥𝐧(𝒆𝒕 + 𝒆−𝒕 ) + 𝐶2 .
𝑑𝑢
9. ∫ 1+sin2 𝛼 𝑑𝛼 = |𝑢 = sin 𝛼 , 𝑑𝑢 = cos 𝛼 𝑑𝛼| = ∫ 1+𝑢2 = arctan 𝑢 + 𝐶 = 𝐚𝐫𝐜𝐭𝐚𝐧(𝐬𝐢𝐧 𝜶) + 𝐶.
𝑥
1
𝑑𝑢
1
𝟏
2
𝟐)
10. ∫ 1+𝑥 4 𝑑𝑥 = |𝑢 = 𝑥 2 , 𝑑𝑢 = 2𝑥𝑑𝑥| = ∫ 1+𝑢
+ 𝐶.
2 = 2 arctan 𝑢 + 𝐶 = 𝟐 𝐚𝐫𝐜𝐭𝐚𝐧(𝒙
𝑑𝑥
1
1
𝑥2
1
1
1
1
11. ∫ 𝑥 arctan 𝑥 𝑑𝑥 = |𝑢 = arctan 𝑥 , 𝑑𝑢 = 1+𝑥 2 , 𝑑𝑣 = 𝑥𝑑𝑥, 𝑣 = 2 𝑥 2 | = 2 𝑥 2 arctan 𝑥 − 2 ∫ 1+𝑥 2 𝑑𝑥 = 2 𝑥 2 arctan 𝑥 − 2 ∫ (1 − 1+𝑥 2 ) 𝑑𝑥 =
1
1
𝟏
𝟏
= 𝑥 2 arctan 𝑥 − (𝑥 − arctan 𝑥) + 𝐶 = (𝒙𝟐 + 𝟏) 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙 − 𝒙 + 𝑪.
2
2
𝟐
𝟐
12. ∫ sin3 𝜑 𝑑𝜑 = |𝑢 = cos 𝜑 , 𝑑𝑢 = − sin 𝜑 𝑑𝜑, sin2 𝜑 = 1 − cos2 𝜑 = 1 − 𝑢2 | =
1
𝟏
= − ∫(1 − 𝑢2 )𝑑𝑢 = −𝑢 + 𝑢3 + 𝐶 = − 𝐜𝐨𝐬 𝝋 + 𝐜𝐨𝐬𝟑 𝝋 + 𝐶.
3
𝟑
𝟏
13. ∫ sec 4 𝑡 𝑑𝑡 = ∫ sec 2 𝑡 (1 + tan2 𝑡)𝑑𝑡 = ∫ sec 2 𝑡 𝑑𝑡 + ∫ sec 2 𝑡 tan2 𝑡 𝑑𝑡 = tan 𝑡 + ∫ tan2 𝑡 𝑑(tan 𝑡) = 𝐭𝐚𝐧 𝒕 + 𝟑 𝐭𝐚𝐧𝟑 𝒕 + 𝐶.
14. ∫
tan(√𝑧)
√𝑧
𝑑𝑧
𝑑𝑧 = |𝑢 = √𝑧, 𝑑𝑢 = 2 𝑧| = 2 ∫ tan 𝑢 𝑑𝑢 = −2 ln|cos 𝑢| + 𝐶 = −𝟐 𝐥𝐧|𝐜𝐨𝐬(√𝒛)| + 𝐶.
√
𝑑𝑢
1
1
𝟏
15. ∫ 52𝑥 𝑑𝑥 = ∫ 𝑒 2𝑥 ln 5 𝑑𝑥 = |𝑢 = 2𝑥 ln 5 , 𝑑𝑢 = 2 ln 5 𝑑𝑥| = ∫ 𝑒 𝑢 2 ln 5 = 2 ln 5 𝑒 𝑢 + 𝐶 = 2 ln 5 𝑒 2𝑥 ln 5 + 𝐶 = 𝟐 𝐥𝐧 𝟓 𝟓𝟐𝒙 + 𝐶.
1
1
𝟏
16. ∫ 𝑒 3𝑥 sec(𝑒 3𝑥 ) 𝑑𝑥 = |𝑢 = 𝑒 3𝑥 , 𝑑𝑢 = 3𝑒 3𝑥 𝑑𝑥| = 3 ∫ sec 𝑢 𝑑𝑢 = 3 ln|tan 𝑢 + sec 𝑢| + 𝐶 = 𝟑 𝐥𝐧|𝐭𝐚𝐧(𝒆𝟑𝒙 ) + 𝐬𝐞𝐜(𝒆𝟑𝒙 )| + 𝐶.
𝑑𝑥
𝑑𝑥
17. ∫ 1+ 𝑥 = |𝑢 = 1 + √𝑥, 𝑑𝑢 = 2 𝑥 , 𝑑𝑥 = 2√𝑥𝑑𝑢 = 2(𝑢 − 1)𝑑𝑢| = ∫
√
√
2(𝑢−1)𝑑𝑢
𝑢
1
= 2 ∫ (1 − 𝑢) 𝑑𝑢 = 2(𝑢 − ln|𝑢|) + 𝐶 =
= 2(1 + √𝑥 − ln(1 + √𝑥)) + 𝐶 = 𝟐(√𝒙 − 𝐥𝐧(𝟏 + √𝒙)) +...