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Title: Integration of quadratic polynomials
Description: This note contains worked examples of integrals containing quadratic polynomials

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[Last Name] 1
INTEGRATION
INTEGRALS CONTAINING QUADRATIC POLYNOMIALS (SPECIAL CASES)
CASE 1
𝟏

Many Integrals in these general forms, ∫ (π’‚π’™πŸ+𝒃𝒙+𝒄) 𝒅𝒙,
∫(

𝟏

) 𝒅𝒙, 𝒄𝒂𝒏 𝒃𝒆 π’”π’Šπ’Žπ’‘π’π’Šπ’‡π’Šπ’†π’… π’ƒπ’š 𝒕𝒉𝒆 𝒑𝒓𝒐𝒄𝒆𝒔𝒔 𝒐𝒇 π’„π’π’Žπ’‘π’π’†π’•π’Šπ’π’ˆ 𝒕𝒉𝒆 𝒔𝒒𝒖𝒂𝒓𝒆
...

𝟏

Worked Example 1: Find ∫ (πŸ—π’™πŸ+πŸ”π’™+πŸ“) 𝒅𝒙
Completing the square
9xΒ²+6x+5
πŸ— (π’™πŸ +

πŸ”π’™

πŸ— (π’™πŸ +

πŸπ’™

πŸ—

πŸ‘

)+πŸ“
)+πŸ“

𝟏 𝟐

𝟏

πŸ— ((𝒙 + πŸ‘) βˆ’ πŸ—) + πŸ“
𝟏 𝟐

πŸ— (𝒙 + ) βˆ’ 𝟏 + πŸ“
πŸ‘

𝟏 𝟐

πŸ— (𝒙 + πŸ‘) + πŸ’
𝟏 𝟐

πŸ‘πŸ (𝒙 + ) + πŸ’
πŸ‘

𝟏

𝟐

(πŸ‘ (𝒙 + πŸ‘)) + πŸ’
(πŸ‘π’™ + 𝟏)𝟐 + πŸ’
Re-writing the Integral

[Last Name] 2
𝟏

∫ ((πŸ‘π’™+𝟏)𝟐+πŸ’) 𝒅𝒙
Applying the substitution, U=3x+1
𝒅𝒖
𝒅𝒙

= πŸ‘

Making β€œdx” the subject
𝒅𝒙 =

𝒅𝒖
πŸ‘

𝟏

= πŸ‘ 𝒅𝒖

Substituting the value of β€œdx” into the Integral and re-writing the Integral in terms of U
𝟏

𝟏

∫ (π’–πŸ+πŸ’)
...
𝟐 𝒅𝒖

Pulling out Β½
𝟏
𝟐

∫(

𝟏
βˆšπŸπŸ“βˆ’π’–πŸ

) 𝒅𝒖

Integrating by trigonometric substitution
Recall, ∫ (

𝟏
βˆšπ’‚πŸ βˆ’π’™πŸ

𝒙

) 𝒅𝒙 = π’”π’Šπ’βˆ’πŸ (𝒂) + 𝒄

Re-writing the Integral in the right pattern for the trigonometric substitution
𝟏
𝟐

∫(

𝟏
βˆšπŸ“πŸ βˆ’π’–πŸ

) 𝒅𝒖

x= u
a=5
𝟏
𝟐

𝒖

π’”π’Šπ’βˆ’πŸ (𝒂) + 𝒄

But, u= 2x-4
∫(

𝟏

𝟏

βˆšπŸ—+πŸπŸ”π’™βˆ’πŸ’π’™πŸ

) 𝒅𝒙 = 𝟐 π’”π’Šπ’βˆ’πŸ (

πŸπ’™βˆ’πŸ’
πŸ“

)+𝒄

CASE 2
Integrals in the form ∫ (
𝒅
𝒅𝒙

𝑨𝑿+𝑩
𝒏
(π’‚π’™πŸ +𝒃𝒙+𝒄)

) 𝒅𝒙, π’˜π’‰π’†π’“π’† n π’Šπ’” 𝒂 π’‘π’π’”π’Šπ’•π’Šπ’—π’† π’Šπ’π’•π’†π’ˆπ’†π’“ 𝒂𝒏𝒅 π’˜π’‰π’†π’“π’†

(π’‚π’™πŸ + 𝒃𝒙 + 𝒄) β‰  𝑨𝒙 + 𝑩, 𝒄𝒂𝒏 𝒃𝒆 𝒔𝒐𝒍𝒗𝒆𝒅 π’ƒπ’š π’”π’‘π’π’Šπ’•π’•π’Šπ’π’ˆ π’Šπ’• π’Šπ’π’•π’ π’”π’Šπ’Žπ’‘π’π’†π’“ π’Šπ’π’•π’†π’ˆπ’“π’‚π’π’”

that are in standard form
...


[Last Name] 6
We are going to re-write 2x+3 as a constant multiple of 18x+6
2x+3=P(18x+6)+Q
2x+3=18Px+6P+Q
Equating corresponding coefficients
2=18P……(I)
6P+Q=3…
...
πŸ‘ 𝒅𝒖
πŸ‘
𝟏

πŸ•

π‘·π’–π’π’π’Šπ’π’ˆ 𝒐𝒖𝒕 πŸ‘ 𝒂𝒏𝒅 π’Žπ’–π’π’•π’Šπ’‘π’π’šπ’Šπ’π’ˆ π’Šπ’• π’˜π’Šπ’•π’‰ πŸ‘
πŸ•
πŸ—

∫(

𝟏

) 𝒅𝒖

π’–πŸ +πŸ’

Integrating by trigonometric substitution
𝟏

𝟏

𝒙

Recall, ∫ (π’™πŸ+π’‚πŸ) 𝒅𝒙 = 𝒂 π’•π’‚π’βˆ’πŸ (𝒂) + 𝒄
Re-writing the Integral in the right pattern for the trigonometric substitution
πŸ•
πŸ—

𝟏

∫ (π’–πŸ+𝟐𝟐) 𝒅𝒖

[Last Name] 8
x= u
a=2
πŸ• 𝟏

𝒖

[ π’•π’‚π’βˆ’πŸ ( 𝟐)] + 𝒄
πŸ— 𝟐
But, U=3x+1
Further simplifying
πŸ•
πŸπŸ–

π’•π’‚π’βˆ’πŸ (

πŸ‘π’™+𝟏

) + π’„πŸ

𝟐

πŸ•

πŸ•

πŸ‘
∫ ( πŸ—π’™πŸ+πŸ”π’™+πŸ“
) 𝒅𝒙 = πŸπŸ– π’•π’‚π’βˆ’πŸ (

πŸ‘π’™+𝟏
𝟐

) + π’„πŸ

Therefore,
πŸπ’™+𝟏

𝟏

πŸ•

∫ (πŸ—π’™πŸ+πŸ”π’™+πŸ“) 𝒅𝒙 = πŸ— 𝒍𝒏(πŸ—π’™πŸ + πŸ”π’™ + πŸ“) + πŸπŸ– π’•π’‚π’βˆ’πŸ (

πŸ‘π’™+𝟏
𝟐

)+π‘ͺ

NOTE: π’„πŸ 𝒂𝒏𝒅 π’„πŸ 𝒉𝒂𝒗𝒆 𝒃𝒆𝒆𝒏 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒅 π’˜π’Šπ’•π’‰ 𝒂 π’”π’Šπ’π’ˆπ’π’† 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 π‘ͺ

WORKED EXAMPLE 2: Find ∫ (

(𝒙+𝟏)
𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

𝒅

In this case, 𝒅𝒙 (π’™πŸ + πŸ’π’™ + πŸ“) = πŸπ’™ + πŸ’
We would need to re-write x+1 as a constant multiple of 2x+4
x+1=P(2x+4)+Q
Further simplifying
x+1=2Px+4P+Q
Equating corresponding coefficients
2P=1…
...
(II)
Solving simultaneously

[Last Name] 9
P=Β½
Q= -1
x+1 = P(2x+4)+Q
x+1= Β½(2x+4)-1
Re-writing the Integral
∫(

𝟏
𝟐

( )(πŸπ’™+πŸ’)βˆ’πŸ
𝟐

(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

Splitting into two simpler Integrals
𝟏

∫(

(𝟐)(πŸπ’™+πŸ’)
𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙 βˆ’ ∫ (

𝟏
𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

Start with the first Integral on the left
𝟏

∫(

(𝟐)(πŸπ’™+πŸ’)
𝟐

(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

Pulling out Β½
𝟏
𝟐

∫(

(πŸπ’™+πŸ’)
𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

Applying the substitution U=π’™πŸ + πŸ’π’™ + πŸ“
𝒖 = π’™πŸ + πŸ’π’™ + πŸ“
𝒅𝒖
𝒅𝒙

= πŸπ’™ + πŸ’

Making β€œdx” the subject
𝒅𝒖

𝟏

𝒅𝒙 = πŸπ’™+πŸ’ = πŸπ’™+πŸ’ 𝒅𝒖
Substituting the value of β€œdx” into the Integral and re-writing the Integral in terms of U
𝟏
𝟐

∫(

πŸπ’™+πŸ’

𝟏

π‘ΌπŸ

πŸπ’™+πŸ’

)
...


𝟐 βˆ’πŸ+𝟏
𝟏

βˆ’ πŸπ‘Ό
But, U= xΒ²+4x+5
𝟏

∫(

(𝟐)(πŸπ’™+πŸ’)

𝟏

𝟐

(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙 = βˆ’ 𝟐(π’™πŸ+πŸ’π’™+πŸ“) + π’„πŸ

Now, solving the second integral
βˆ’βˆ« (

(𝟏)
𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

) 𝒅𝒙

Completing the square
xΒ²+4x+5
(xΒ²+4x)+5
((x+2)Β²-4)+5
(x+2)Β²-4+5
(x+2)Β²+1
Re-writing the Integral
∫(

𝟏
𝟐

((𝒙+𝟐)𝟐 +𝟏)

) 𝒅𝒙

Applying the substitution U=x+2
U=x+2
𝒅𝒖
𝒅𝒙

=𝟏

Making β€œdx” the subject
𝒅𝒙 = 𝒅𝒖

[Last Name] 11
Substituting the value of β€œdx” into the Integral and re-writing the Integral in terms of U
∫(

𝟏
𝟐
(π’–πŸ +𝟏)

) 𝒅𝒖

Integrating by trigonometric substitution
Recall,
∫(

𝟏

𝒙

𝟐
(π’‚πŸ +π’™πŸ )

𝟏

𝒙

) 𝒅𝒙 = πŸπ’‚πŸ (π’‚πŸ+π’™πŸ) + πŸπ’‚πŸ‘ π’•π’‚π’βˆ’πŸ (𝒂) + 𝒄

a=1
x=u
Substituting the values of β€œa” and β€œx” it becomes;
βˆ’βˆ« (

𝟏

𝒙+𝟐

𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

Therefore, ∫ (

𝟏

) 𝒅𝒙 = βˆ’ 𝟐(π’™πŸ+πŸ’π’™+πŸ“) βˆ’ 𝟐 𝒕𝒂𝒏(𝒙 + 𝟐) + π‘ͺ𝟐
𝒙+𝟏

𝟐
(π’™πŸ +πŸ’π’™+πŸ“)

𝟏

𝒙+𝟐

𝟏

) 𝒅𝒙 = βˆ’ 𝟐(π’™πŸ+πŸ’π’™+πŸ“) βˆ’ 𝟐(π’™πŸ+πŸ’π’™+πŸ“) βˆ’ 𝟐 𝒕𝒂𝒏(𝒙 + 𝟐) + 𝒄
βˆ’π’™βˆ’πŸ‘

𝟏

= 𝟐(π’™πŸ+πŸ’π’™+πŸ“) βˆ’ 𝟐 𝒕𝒂𝒏(𝒙 + 𝟐) + π‘ͺ
NOTE: π’„πŸ 𝒂𝒏𝒅 π’„πŸ 𝒉𝒂𝒗𝒆 𝒃𝒆𝒆𝒏 𝒓𝒆𝒑𝒍𝒂𝒄𝒆𝒅 π’˜π’Šπ’•π’‰ 𝒂 π’”π’Šπ’π’ˆπ’π’† 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 π‘ͺ
Title: Integration of quadratic polynomials
Description: This note contains worked examples of integrals containing quadratic polynomials