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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 22 Find (i) โˆซ1โ–’๐‘’^๐‘ฅ (tan^(โˆ’1)โก๐‘ฅ+ 1/(1 + ๐‘ฅ^2 )) ๐‘‘๐‘ฅ โˆซ1โ–’ใ€–๐‘’^๐‘ฅ (tan^(โˆ’1)โก๐‘ฅ+1/(1 + ๐‘ฅ^2 ))๐‘‘๐‘ฅใ€— It is of the form โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [๐‘“(๐‘ฅ)+๐‘“^โ€ฒ (๐‘ฅ)] ใ€— ๐‘‘๐‘ฅ=๐‘’^๐‘ฅ ๐‘“(๐‘ฅ)+๐ถ Where ๐‘“(๐‘ฅ)=tan^(โˆ’1)โก๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)= 1/(1 + ๐‘ฅ^2 ) So, our equation becomes โˆซ1โ–’ใ€–๐‘’^๐‘ฅ (tan^(โˆ’1)โก๐‘ฅ+1/(1 + ๐‘ฅ^2 ))๐‘‘๐‘ฅใ€—=๐’†^๐’™ ใ€–๐ญ๐š๐งใ€—^(โˆ’๐Ÿ)โกใ€–๐’™+๐‘ชใ€— Example 22 Find (ii) โˆซ1โ–’((๐‘ฅ^2 + 1) ๐‘’^๐‘ฅ)/(๐‘ฅ + 1)^2 ๐‘‘๐‘ฅ โˆซ1โ–’ใ€–(๐‘ฅ^2 + 1)/(๐‘ฅ + 1)^2 .๐‘’^๐‘ฅ ๐‘‘๐‘ฅใ€— Adding and subtracting 1 in numerator =โˆซ1โ–’ใ€–(๐‘ฅ^2+ 1 + 1 โˆ’ 1)/(๐‘ฅ + 1)^2 .๐‘’^๐‘ฅ .๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–(๐‘ฅ^2 โˆ’ 1 + 1 + 1)/(๐‘ฅ + 1)^2 .๐‘’^๐‘ฅ .๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–[(๐‘ฅ^2 โˆ’ 1)/(๐‘ฅ + 1)^2 +2/(๐‘ฅ + 1)^2 ] ๐‘’^๐‘ฅ ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [(๐‘ฅ^2 โˆ’ (1)^2)/(๐‘ฅ + 1)^2 +2/(๐‘ฅ + 1)^2 ]๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [(๐‘ฅ โˆ’ 1)(๐‘ฅ + 1)/(๐‘ฅ + 1)^2 +2/(๐‘ฅ + 1)^2 ]๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [(๐‘ฅ โˆ’ 1)/(๐‘ฅ + 1)+2/(๐‘ฅ + 1)^2 ]๐‘‘๐‘ฅใ€— It is of form โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [๐‘“(๐‘ฅ)+๐‘“^โ€ฒ (๐‘ฅ)] ใ€— ๐‘‘๐‘ฅ=๐‘’^๐‘ฅ ๐‘“(๐‘ฅ)+๐ถ Where ๐‘“(๐‘ฅ)=(๐‘ฅ โˆ’ 1)/(๐‘ฅ + 1) ๐‘“^โ€ฒ (๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ [(๐‘ฅ โˆ’ 1)/(๐‘ฅ + 1)] ๐‘“^โ€ฒ (๐‘ฅ)=(1.(๐‘ฅ + 1) โˆ’1 (๐‘ฅ โˆ’ 1))/(๐‘ฅ + 1)^2 =(๐‘ฅ + 1 โˆ’ ๐‘ฅ + 1)/(๐‘ฅ + 1)^2 =2/(๐‘ฅ + 1)^2 Thus, our equation becomes โˆซ1โ–’ใ€–(๐‘ฅ^2 + 1)/(๐‘ฅ + 1)^2 .๐‘’^๐‘ฅ=โˆซ1โ–’ใ€–๐‘’^๐‘ฅ [(๐‘ฅ โˆ’ 1)/(๐‘ฅ + 1)+2/(๐‘ฅ + 1)^2 ]๐‘‘๐‘ฅใ€—ใ€— =๐‘’^๐‘ฅ [(๐‘ฅ โˆ’ 1)/(๐‘ฅ + 1)]+๐ถ =(๐’™ โˆ’ ๐Ÿ)/(๐’™ + ๐Ÿ).๐’†^๐’™+๐‘ช

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.