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TABLE OF ANTIDIFFERENTIATION FORMULAS
General antiderivatives
1
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∫ 𝑎𝑓 (𝑥 ) 𝑑𝑥 = 𝑎 ∫ 𝑓 (𝑥 ) 𝑑𝑥
𝑥 𝑛+1

3
...
∫ 𝑥 𝑛 𝑑𝑥 = ln|𝑥 | + 𝐶, 𝑛 = −1
𝑎𝑛

5
...
∫ 𝑒 𝑛 𝑑𝑥 = 𝑒 𝑛 + 𝐶

Antidifferentiation formulas of trigonometric functions
7
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∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
9
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∫ csc2 𝑥 𝑑𝑥 = −cot 𝑥 + 𝐶
11
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∫ csc 𝑥 cot 𝑥 𝑑𝑥 = −csc 𝑥 + 𝐶
13
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∫ tan 𝑥 𝑑𝑥 = ln|sec 𝑥 | + 𝐶
15
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∫ csc 𝑥 𝑑𝑥 = ln|csc 𝑥 − cot 𝑥 | + 𝐶
Antidifferentiation formulas yielding inverse trigonometric functions
1

17
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∫ 𝑎2 +𝑥2 𝑑𝑥 = 𝑎 tan−1 (𝑎) + 𝐶, 𝑎 ≠ 0
19
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∫ sinh 𝑥 𝑑𝑥 = cosh 𝑥 + 𝐶
21
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∫ sech2 𝑥 𝑑𝑥 = tanh 𝑥 + 𝐶
23
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∫ sech 𝑥 tanh 𝑥 𝑑𝑥 = −sech 𝑥 + 𝐶
25
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∫ coth 𝑥 𝑑𝑥 = ln|sinh 𝑥 | + 𝐶
27
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∫ sech 𝑥 𝑑𝑥 = 2 tan−1 𝑒 𝑥 + 𝐶 = tan−1 (sinh 𝑥 ) + 𝐶
29
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Antiderivatives are not unique
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𝐹 (𝑥 ) + 𝐶 is the general antiderivative of 𝑓, while each antiderivative is a particular
antiderivative of 𝑓
...

Example: Given that 𝐹 ′ (𝑥 ) = 2𝑥 3 and that 𝐹 (4) = 16, what is 𝐹(𝑥)?
First, identify the general antiderivative of 𝐹(𝑥)
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𝐹 (4) = 16
(4)4
+ 𝐶 = 16
4
𝐶 = 16 − 64 = −48
Thus, the particular antiderivative is 𝑭(𝒙) =

INTEGRATION BY SUBSTITUTION

𝒙𝟒
𝟐

− 𝟒𝟖
...
The challenging
part with substitution is with finding the appropriate value of 𝑢
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Suppose 𝐹(𝑥) is an antiderivative of 𝑓(𝑥), then
𝐹(𝑔(𝑥 )) is an antiderivative of 𝑓(𝑔(𝑥 )) ∙ 𝑔′(𝑥)
...
Thus,
∫ 𝑓 (𝑢)𝑑𝑢 = 𝐹 (𝑢) + 𝐶
Candidates for 𝑢 include the following:
a
...

c
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The denominator of the function or a part thereof
Non-integrable functions
The exponent or a part thereof
The insides of radicals
𝑥3

Example: ∫ (𝑥4 +4)5 𝑑𝑥
The function is non-integrable as of the moment
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The
function satisfies candidate (a)
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Thus, 𝑑𝑢 = 4𝑥 3 𝑑𝑥
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e
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Thus,
∫

𝑥3
1 1
1 1
𝑑𝑥 = ∫ 5 ∙ 𝑑𝑢 = ∫ 5 𝑑𝑢
4
5
(𝑥 + 4)
𝑢 4
4 𝑢

The substituted function can now be integrated
...

1 1
1 𝑢−4
1
∫ 5 𝑑𝑢 = (
)+𝐶 =−
+𝐶
4 𝑢
4 −4
16𝑢4
We then revert 𝑢 back to 𝑥
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Thus, 𝑑𝑢 = cos 𝑥 𝑑𝑥
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We then perform a second substitution
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Thus, 𝑑𝑣 = 3𝑢2 𝑑𝑢
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This can be resolved by splitting 𝑢8 such that it can be utilized by
𝑑𝑣 and so that it can be expressed as 𝑣
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Solving for 𝑢3 yields 𝑢3 = 𝑣 − 3

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