Example 9 - Chapter 7 Class 12 Integrals - Part 3

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Example 9 - Chapter 7 Class 12 Integrals - Part 4

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

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Example 9 Find the following integrals: (ii) โˆซ1โ–’๐‘‘๐‘ฅ/(ใ€–3๐‘ฅใ€—^2โˆ’13๐‘ฅ + 10) โˆซ1โ–’๐‘‘๐‘ฅ/(ใ€–3๐‘ฅใ€—^2 โˆ’ 13๐‘ฅ + 10) Solving denominator ใ€–3๐‘ฅใ€—^2+13๐‘ฅโˆ’10 =3(๐‘ฅ^2+13/3 ๐‘ฅ โˆ’10/3) =3(๐‘ฅ^2+2. ๐‘ฅร— 13/6 โˆ’10/3) Adding and subtracting (13/6)^2 =3(๐‘ฅ^2+2. ๐‘ฅร— 13/6+(13/6)^2โˆ’10/3โˆ’(13/6)^2 ) =3((๐‘ฅ+13/6)^2โˆ’10/3โˆ’(169/36)) =3((๐‘ฅ+13/6)^2โˆ’(10/3 +169/36)) =3((๐‘ฅ+13/6)^2โˆ’((120 +169)/36 )) =3((๐‘ฅ+13/6)^2โˆ’289/36) =3((๐‘ฅ+13/6)^2โˆ’(17/6)^2 ) Hence, our equation becomes โˆซ1โ–’๐‘‘๐‘ฅ/(ใ€–3๐‘ฅใ€—^2 โˆ’ 13๐‘ฅ + 10) = 1/3 โˆซ1โ–’๐‘‘๐‘ฅ/((๐‘ฅ + 13/6)^2โˆ’ (17/6)^2 ) It is of form โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/(๐‘ฅ^2 โˆ’ ๐‘Ž^2 )=1/2๐‘Ž ๐‘™๐‘œ๐‘”|(๐‘ฅ โˆ’ ๐‘Ž)/(๐‘ฅ + ๐‘Ž)|+๐ถ1ใ€— Replacing ๐‘ฅ by (๐‘ฅ+13/6)๐‘Ž๐‘›๐‘‘ ๐‘Ž ๐‘๐‘ฆ 17/6, = 1/3 ร— 1/2(17/6) ร—logโก|(๐‘ฅ + 13/6 โˆ’ 17/6)/(๐‘ฅ+ 13/6 + 17/6)| + C = 1/3 ร— 6/2(17) ร—logโก|((6๐‘ฅ + 13 โˆ’ 17)/6)/((6๐‘ฅ +13 + 17)/6)| + C = 1/17 logโก|(6๐‘ฅ โˆ’ 4)/(6๐‘ฅ + 30)| + C = 1/17 logโก|(2(3๐‘ฅ โˆ’ 2))/(6(๐‘ฅ + 5))|+ C = 1/17 logโก|( (3๐‘ฅ โˆ’ 2))/(3(๐‘ฅ + 5))|+ C = 1/17 logโก|( (3๐‘ฅ โˆ’ 2))/((๐‘ฅ + 5))|โˆ’1/17 logโก3 + C = ๐Ÿ/๐Ÿ๐Ÿ• ๐’๐’๐’ˆโก|( (๐Ÿ‘๐’™ โˆ’ ๐Ÿ))/((๐’™ + ๐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 10 years. He provides courses for Maths and Science at Teachoo.