[Last Name] 1
INTEGRATION
INTEGRALS CONTAINING QUADRATIC POLYNOMIALS (SPECIAL CASES)
CASE 1
𝟏
Many Integrals in these general forms, ∫ (𝒂𝒙𝟐+𝒃𝒙+𝒄) 𝒅𝒙,
𝟏
∫( ) 𝒅𝒙, 𝒄𝒂𝒏 𝒃𝒆 𝒔𝒊𝒎𝒑𝒍𝒊𝒇𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒑𝒓𝒐𝒄𝒆𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒒𝒖𝒂𝒓𝒆.
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
This method is applied in order to transform such Integrals to their standard form, giving
room for the Appropriate integration technique to be easily used.
𝟏
Worked Example 1: Find ∫ (𝟗𝒙𝟐+𝟔𝒙+𝟓) 𝒅𝒙
Completing the square
9x²+6x+5
𝟔𝒙
𝟗 (𝒙𝟐 + )+𝟓
𝟗
𝟐𝒙
𝟗 (𝒙𝟐 + )+𝟓
𝟑
𝟏 𝟐 𝟏
𝟗 ((𝒙 + 𝟑) − 𝟗) + 𝟓
𝟏 𝟐
𝟗 (𝒙 + ) − 𝟏 + 𝟓
𝟑
𝟏 𝟐
𝟗 (𝒙 + 𝟑) + 𝟒
𝟏 𝟐
𝟑𝟐 (𝒙 + ) + 𝟒
𝟑
𝟐
𝟏
(𝟑 (𝒙 + 𝟑)) + 𝟒
(𝟑𝒙 + 𝟏)𝟐 + 𝟒
Re-writing the Integral
, [Last Name] 2
𝟏
∫ ((𝟑𝒙+𝟏)𝟐+𝟒) 𝒅𝒙
Applying the substitution, U=3x+1
𝒅𝒖
= 𝟑
𝒅𝒙
Making “dx” the subject
𝒅𝒖 𝟏
𝒅𝒙 = = 𝟑 𝒅𝒖
𝟑
Substituting the value of “dx” into the Integral and re-writing the Integral in terms of U
𝟏 𝟏
∫ (𝒖𝟐+𝟒) . 𝟑 𝒅𝒖
Pulling out ⅓
𝟏 𝟏
∫ (𝒖𝟐+𝟒) 𝒅𝒖
𝟑
Now, integrating by trigonometric substitution
𝟏 𝟏 𝒙
Recall, ∫ (𝒙𝟐+𝒂𝟐) 𝒅𝒙 = 𝒂 𝒕𝒂𝒏−𝟏 (𝒂) + 𝒄
Re-writing the Integral in the required pattern for the trigonometric substitution
𝟏 𝟏
∫ (𝒖𝟐+𝟐𝟐) 𝒅𝒖
𝟑
Where x=u, and a=2
We would then have;
𝟏 𝟏 𝒖
[ 𝒕𝒂𝒏−𝟏 ( 𝟐)]
𝟑 𝟐
Removing the bracket
𝟏 𝒖
𝒕𝒂𝒏−𝟏 ( 𝟐) + 𝒄
𝟔
Recall, U=3x+1
𝟏 𝟏 𝟑𝒙+𝟏
∫ (𝟗𝒙𝟐+𝟔𝒙+𝟓) 𝒅𝒙 = 𝟔 𝒕𝒂𝒏−𝟏 ( )+𝒄
𝟐
INTEGRATION
INTEGRALS CONTAINING QUADRATIC POLYNOMIALS (SPECIAL CASES)
CASE 1
𝟏
Many Integrals in these general forms, ∫ (𝒂𝒙𝟐+𝒃𝒙+𝒄) 𝒅𝒙,
𝟏
∫( ) 𝒅𝒙, 𝒄𝒂𝒏 𝒃𝒆 𝒔𝒊𝒎𝒑𝒍𝒊𝒇𝒊𝒆𝒅 𝒃𝒚 𝒕𝒉𝒆 𝒑𝒓𝒐𝒄𝒆𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒍𝒆𝒕𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒒𝒖𝒂𝒓𝒆.
√𝒂𝒙𝟐 +𝒃𝒙+𝒄
This method is applied in order to transform such Integrals to their standard form, giving
room for the Appropriate integration technique to be easily used.
𝟏
Worked Example 1: Find ∫ (𝟗𝒙𝟐+𝟔𝒙+𝟓) 𝒅𝒙
Completing the square
9x²+6x+5
𝟔𝒙
𝟗 (𝒙𝟐 + )+𝟓
𝟗
𝟐𝒙
𝟗 (𝒙𝟐 + )+𝟓
𝟑
𝟏 𝟐 𝟏
𝟗 ((𝒙 + 𝟑) − 𝟗) + 𝟓
𝟏 𝟐
𝟗 (𝒙 + ) − 𝟏 + 𝟓
𝟑
𝟏 𝟐
𝟗 (𝒙 + 𝟑) + 𝟒
𝟏 𝟐
𝟑𝟐 (𝒙 + ) + 𝟒
𝟑
𝟐
𝟏
(𝟑 (𝒙 + 𝟑)) + 𝟒
(𝟑𝒙 + 𝟏)𝟐 + 𝟒
Re-writing the Integral
, [Last Name] 2
𝟏
∫ ((𝟑𝒙+𝟏)𝟐+𝟒) 𝒅𝒙
Applying the substitution, U=3x+1
𝒅𝒖
= 𝟑
𝒅𝒙
Making “dx” the subject
𝒅𝒖 𝟏
𝒅𝒙 = = 𝟑 𝒅𝒖
𝟑
Substituting the value of “dx” into the Integral and re-writing the Integral in terms of U
𝟏 𝟏
∫ (𝒖𝟐+𝟒) . 𝟑 𝒅𝒖
Pulling out ⅓
𝟏 𝟏
∫ (𝒖𝟐+𝟒) 𝒅𝒖
𝟑
Now, integrating by trigonometric substitution
𝟏 𝟏 𝒙
Recall, ∫ (𝒙𝟐+𝒂𝟐) 𝒅𝒙 = 𝒂 𝒕𝒂𝒏−𝟏 (𝒂) + 𝒄
Re-writing the Integral in the required pattern for the trigonometric substitution
𝟏 𝟏
∫ (𝒖𝟐+𝟐𝟐) 𝒅𝒖
𝟑
Where x=u, and a=2
We would then have;
𝟏 𝟏 𝒖
[ 𝒕𝒂𝒏−𝟏 ( 𝟐)]
𝟑 𝟐
Removing the bracket
𝟏 𝒖
𝒕𝒂𝒏−𝟏 ( 𝟐) + 𝒄
𝟔
Recall, U=3x+1
𝟏 𝟏 𝟑𝒙+𝟏
∫ (𝟗𝒙𝟐+𝟔𝒙+𝟓) 𝒅𝒙 = 𝟔 𝒕𝒂𝒏−𝟏 ( )+𝒄
𝟐
