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Example 9 - Chapter 7 Class 12 Integrals - Part 3

Example 9 - Chapter 7 Class 12 Integrals - Part 4
Example 9 - Chapter 7 Class 12 Integrals - Part 5
Example 9 - Chapter 7 Class 12 Integrals - Part 6


Transcript

Example 9 Find the following integrals: (ii) ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2βˆ’13π‘₯ + 10) ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2 βˆ’ 13π‘₯ + 10) Solving denominator γ€–3π‘₯γ€—^2+13π‘₯βˆ’10 =3(π‘₯^2+13/3 π‘₯ βˆ’10/3) =3(π‘₯^2+2. π‘₯Γ— 13/6 βˆ’10/3) Adding and subtracting (13/6)^2 =3(π‘₯^2+2. π‘₯Γ— 13/6+(13/6)^2βˆ’10/3βˆ’(13/6)^2 ) =3((π‘₯+13/6)^2βˆ’10/3βˆ’(169/36)) =3((π‘₯+13/6)^2βˆ’(10/3 +169/36)) =3((π‘₯+13/6)^2βˆ’((120 +169)/36 )) =3((π‘₯+13/6)^2βˆ’289/36) =3((π‘₯+13/6)^2βˆ’(17/6)^2 ) Hence, our equation becomes ∫1▒𝑑π‘₯/(γ€–3π‘₯γ€—^2 βˆ’ 13π‘₯ + 10) = 1/3 ∫1▒𝑑π‘₯/((π‘₯ + 13/6)^2βˆ’ (17/6)^2 ) It is of form ∫1▒〖𝑑π‘₯/(π‘₯^2 βˆ’ π‘Ž^2 )=1/2π‘Ž π‘™π‘œπ‘”|(π‘₯ βˆ’ π‘Ž)/(π‘₯ + π‘Ž)|+𝐢1γ€— Replacing π‘₯ by (π‘₯+13/6)π‘Žπ‘›π‘‘ π‘Ž 𝑏𝑦 17/6, = 1/3 Γ— 1/2(17/6) Γ—log⁑|(π‘₯ + 13/6 βˆ’ 17/6)/(π‘₯+ 13/6 + 17/6)| + C = 1/3 Γ— 6/2(17) Γ—log⁑|((6π‘₯ + 13 βˆ’ 17)/6)/((6π‘₯ +13 + 17)/6)| + C = 1/17 log⁑|(6π‘₯ βˆ’ 4)/(6π‘₯ + 30)| + C = 1/17 log⁑|(2(3π‘₯ βˆ’ 2))/(6(π‘₯ + 5))|+ C = 1/17 log⁑|( (3π‘₯ βˆ’ 2))/(3(π‘₯ + 5))|+ C = 1/17 log⁑|( (3π‘₯ βˆ’ 2))/((π‘₯ + 5))|βˆ’1/17 log⁑3 + C = 𝟏/πŸπŸ• π’π’π’ˆβ‘|( (πŸ‘π’™ βˆ’ 𝟐))/((𝒙 + πŸ“))|+ C1

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.