Academic Skills Advice
Integration Summary
Notation
𝑑𝑦
You might be given 𝑑𝑥 and asked to find 𝑦, so you will need to integrate.
Or the question might use the integration notation as follows:
∫
Means:
(6𝑥 2 − 5𝑥 + 7 )𝑑𝑥
with respect to 𝑥
Integrate the following
Basic integration of Polynomials
e.g.
∫
𝑎𝑥 𝑛 𝑑𝑥 =
∫
𝑥 2 − 2𝑥 + 7 𝑑𝑥
=
𝑥3
3
𝑎
𝑥 𝑛+1
𝑛+1
Add 1 to the power
then divide the
expression by the
new power
+𝑐
− 𝑥 2 + 7𝑥 + 𝑐
(Remember to include “+𝑐” in case there was a constant in the original expression).
Area under a curve (using limits)
To find the area under a curve between 2 points, integrate the curve, substitute both points
in separately then subtract the lower limit from the upper.
Upper limit
e.g.
∫
5
6𝑥 2 − 2𝑥 𝑑𝑥
=
[
6𝑥 3
3
−
2𝑥 2
2
]
5
5
=
2
[2𝑥 3 − 𝑥 2 ] 2
2
Lower limit
“Under the curve”
means between the
curve and the 𝑥-axis.
Now do 2 boxes substituting in the limits:
[𝑡𝑜𝑝 𝑙𝑖𝑚𝑖𝑡] − [𝑏𝑜𝑡𝑡𝑜𝑚 𝑙𝑖𝑚𝑖𝑡]
= [2(5)3 − (5)2 ] − [2(2)3 − (2)2 ] = 225 – 12 = 213
We have found that the area under the curve between 𝑥 = 2 and 𝑥 = 5 is 213.
(We didn’t need “+𝑐” in each box because it would have cancelled out.)
© H Jackson 2010 / 2015 / Academic Skills
1
Reverse chain rule
For some integration problems (e.g. when you have a function of a function) you will need
to use the reverse chain rule. This is because the original function was differentiated using
the chain rule and so in order to integrate we need to “undo” the chain rule.
Remember:
Chain rule says:
So reverse chain rule says:
e.g.
e.g.
e.g.
∫
∫
𝑒 7𝑥 𝑑𝑥
𝑒 7𝑥
=
7
sin(5𝑥) 𝑑𝑥
∫
differentiate the inside function and multiply
differentiate the inside function and divide
= −
cos(5𝑥)
Remember:
You will divide by the
new power and the
differential of the inside.
5
(4𝑥−2)7
Outside
(4𝑥 − 2)6 𝑑𝑥 =
(add one to power then
divide by new power)
7 (4)
Inside
(we also have to divide by 4 because
when we differentiate (4𝑥 − 2) we get 4)
Reverse chain rule (some patterns to look out for)
The following patterns always use the reverse chain rule to integrate:
∫
𝑓
′ (𝑥)(𝑓(𝑥))𝑛
𝑑𝑥
=
1
𝑛+1
(𝑓(𝑥))
𝑛+1
+ 𝑐
Just ignore the 1st bit and integrate the
‘outside’ of the 2nd bit (i.e. add one to the
power then divide by the new power).
∫
𝑓′ (𝑥)
𝑓(𝑥)
𝑑𝑥
Integrates to ln of the denominator.
= ln(𝑓(𝑥)) + 𝑐
These rules may look complicated but they are exactly the same as the previous examples,
we just have to check that we have the correct pattern first.
Step 1: Check that the pattern is correct.
Step 2: Integrate as usual, using the reverse chain rule.
e.g.
∫
𝟑𝒙𝟐 (𝒙𝟑 − 𝟓)𝟔 𝒅𝒙
Step 1: the pattern is correct as we have a function inside the bracket and its
differential next to it.
Add 1 to power
Step 2:
=
3𝑥 2 (𝑥 3 −5)7
7 (3𝑥 2 )
Divide by new power
© H Jackson 2010 / 2015 / Academic Skills
cancels to give:
(𝑥 3 −5)7
7
Divide by
differential of inside
2
e.g.
∫
𝟓
𝟓𝒙+𝟏
𝒅𝒙
Step 1: the pattern is correct as we have a function on the bottom and its differential on
the top.
=
Step 2:
5ln(5𝑥)
cancels to give:
5
ln(5𝑥)
Divide by
differential of inside
The method still works if the 𝑓 ′ (𝑥) part is a factor, or multiple, of 𝑓(𝑥). See the following
examples:
e.g.
∫
𝟔𝒙(𝒙𝟐 + 𝟑)𝟕 𝒅𝒙
Step 1: the pattern is correct as we have a function inside the bracket and a multiple of
its differential next to it.
Add 1 to power
=
Step 2:
6𝑥(𝑥 2 +3)8
8 (2𝑥)
∫
𝟑𝒙
𝟔𝒙𝟐 +𝟖
3(𝑥 2 +8)8
8
Divide by
differential of inside
Divide by new power
e.g.
cancels to give:
𝒅𝒙
Step 1: the pattern is correct as we have a function on the bottom and a factor of its
differential on the top.
Step 2:
=
3𝑥ln(6𝑥 2 +8)
12𝑥
cancels to give:
ln(6𝑥 2 +8)
4
Divide by
differential of inside
© H Jackson 2010 / 2015 / Academic Skills
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By parts
When choosing which
term to call 𝑢 look for:
This is used to integrate a function multiplied by a function.
∫
𝑢
𝑑𝑣
𝑑𝑥
𝑑𝑥
∫
𝑢𝑣 − 𝑣
=
𝑑𝑢
𝑑𝑥
1st – ln(𝑥)
2nd - 𝑥 𝑛
3rd - 𝑒 𝑥
𝑑𝑥
Similar to the product and quotient rule (see differentiation) – write down the 4 bits of
information you need then put them into the formula. Note that your answer has an
integral sign in it so you’ll need to integrate whatever is inside it.
𝑑𝑣
Remember: you are calling one bit 𝑢 and the other bit 𝑑𝑥.
Reverse product Rule
You need to remember the product rule for differentiation – refer to the differentiation
summary if necessary. Once you have spotted this pattern you can just write the answer.
Differentiates to
Differentiates to
∫
𝑢
𝑑𝑣
𝑑𝑥
+ 𝑣
𝑑𝑢
𝑑𝑥
𝑑𝑥
=
𝑢𝑣
e.g.
∫
𝑒𝑥
3
𝑑𝑦
3
3
+ 3𝑥 2 𝑒 𝑥 𝑦 = 𝑒 𝑥 𝑦
𝑑𝑥
Differentiates to
Differentiates to
Integrating factor
𝑑𝑦
Used for functions of the form: 𝑑𝑥 + 𝑝(𝑥)𝑦 = 𝑄(𝑥). The integrating factor will convert the
function into a reverse product rule.
𝒑(𝒙)𝒅𝒙
Integrating factor = 𝒆 ∫
(notice that 𝑝(𝑥) is everything with the 𝑦)
 Find the integrating factor
 Multiply the whole function by the integrating factor
 Solve the left hand side using the reverse product rule.
e.g. Solve:
𝒅𝒚
𝒅𝒙
+ 𝟐𝒙𝒚 = 𝟐𝒆−𝒙
𝟐
Integrating factor = 𝑒 ∫𝑝(𝑥)𝑑𝑥 = 𝑒 ∫2𝑥𝑑𝑥 = 𝒆𝒙
Multiply by the integrating factor:
Tidy up the right hand side:
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝟐
2
2
2
2
2
2
𝑒𝑥 + 2𝑥𝑦𝑒𝑥 = 2𝑒 −𝑥 𝑒𝑥
𝑒𝑥 + 2𝑥𝑦𝑒𝑥 = 2
𝑑𝑥
Now integrate (using the reverse product rule for the LHS) and complete the question.
© H Jackson 2010 / 2015 / Academic Skills
4
Separation of Variables
𝑑𝑦
This is used to integrate a function which consists of 𝑥′𝑠, 𝑦′𝑠 and 𝑑𝑥 ’s mixed up (i.e. of the
𝑑𝑦
form 𝑑𝑥 = 𝑓 (𝑥, 𝑦)). We cannot integrate directly because of the 𝑦 on the right hand side.
To integrate: collect all the 𝑦 bits on one side and all the 𝑥 bits on the other.
𝑑𝑦
e.g.
𝑑𝑥
∫
𝑥2
=
∫
becomes (𝑦 − 3)𝑑𝑦 = 𝑥 2 𝑑𝑥
𝑦−3
The integration symbols are introduced after rearranging.
Substitution
A substitution can be used to simplify a complex integration. For this method you need to
replace any 𝑥 or 𝑑𝑥 with a 𝑢 or 𝑑𝑢 and then integrate as normal.
e.g.
∫
𝑥(𝑥 − 7)5 𝑑𝑥
We can find all of these (and any
others we need) by differentiating
and/or rearranging the 𝑢 = 𝑥 − 7
Let 𝑢 = 𝑥 − 7
𝑑𝑢
∴
𝑑𝑥
=1
so 𝑑𝑢 = 𝑑𝑥
and 𝑥 = 𝑢 + 7
Substitute the above information into the original to produce an integration in terms of 𝑢.
The question now becomes:
∫
∫
(𝑢 + 7)𝑢5 𝑑𝑢
= 𝑢6 + 7𝑢5 𝑑𝑢
=
𝑢7
7
+
7𝑢6
6
+𝑐
=
(𝑥−7)7
© H Jackson 2010 / 2015 / Academic Skills
7
+
7(𝑥−7)6
6
+𝑐
5
Substitution (2 variables & homogenous)
Used when the 𝑥’s and 𝑦’s can’t be separated and when the differential equation is
homogenous. Homogenous means that the total degree in 𝑥 and 𝑦, for each term
involved, is the same (e.g. 𝑥𝑦 is degree 2 and 𝑥 2 is degree 2).
The method is as follows:
1. Substitute 𝒚 = 𝒗𝒙
(where 𝑣 is a function of 𝑥),
2. Differentiate 𝑦 with respect to 𝑥 using the product rule,
3. Substitute everything back into the original function, and cancel where possible so that
you have a function with 𝑥 and 𝑣 which can then be integrated by separating the
variables.
e.g.
𝒅𝒚
𝒅𝒙
=
𝟒𝒚−𝒙
Using the product rule (and implicit differentiation).
𝑦 = 𝑣𝑥
1. Let
2. ∴
𝟑𝒙
𝑑𝑦
𝑑𝑥
(This will be the same every time so you can just
memorise it if you prefer.)
𝑑𝑣
= 𝑣 + 𝑥 𝑑𝑥
𝑑𝑦
3. Replace any 𝑦’s and 𝑑𝑥 in the original:
𝒅𝒚
Original question:
𝒅𝒙
𝑑𝑣
𝑣 + 𝑥 𝑑𝑥 =
Becomes:
𝑑𝑣
Simplify & rearrange:
𝟒𝒚−𝒙
𝟑𝒙
4𝑣𝑥−𝑥
The 𝑥’s will cancel
3𝑥
𝑥 𝑑𝑥 =
4𝑣−1
𝑑𝑣
𝑣−1
𝑥 𝑑𝑥 =
=
3
−𝑣
Combine the fractions:
4𝑣−1−3𝑣
3
3
Now we can separate the variables and solve as normal:
∫
3
𝑣−1
𝑑𝑣 =
∫
1
𝑥
𝑑𝑥
3 ln(𝑣 − 1) = 𝑙𝑛𝑥 + 𝑙𝑛𝐴
ln(𝑣 − 1)3 = ln(𝐴𝑥)
(𝑣 − 1)3 = 𝐴𝑥
Remember that 𝑦 = 𝑣𝑥
∴𝑣=
𝑦
𝑥
𝑦
Don’t forget the constant.
(with logs, 𝑐 = 𝑙𝑛𝐴)
3
(𝑥 − 1) = 𝐴𝑥
𝑦
3
− 1 = √𝐴𝑥
𝑥
3
𝑦 − 𝑥 = √𝐴𝑥 . 𝑥
(𝑦 − 𝑥)3 = 𝐴𝑥 4
© H Jackson 2010 / 2015 / Academic Skills
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