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  1. Chapter 7 Class 12 Integrals
  2. Concept wise

Transcript

Ex 7.9, 19 โˆซ_0^2โ–’(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ Let F(๐‘ฅ)=โˆซ1โ–’ใ€–(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–3/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ ใ€—ใ€— Solving ๐‘ฐ๐Ÿ ๐ผ1=โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— Put ๐‘ฅ^2 + 4=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^2+4)=๐‘‘๐‘ก/๐‘‘๐‘ฅ 2๐‘ฅ+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Therefore, โˆซ1โ–’ใ€–6๐‘ฅ/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ=โˆซ1โ–’ใ€–6๐‘ฅ/๐‘ก ๐‘‘๐‘ก/2๐‘ฅใ€—ใ€— =โˆซ1โ–’ใ€–3/๐‘ก ๐‘‘๐‘กใ€— =3 ๐‘™๐‘œ๐‘”|๐‘ก| =3 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+4| (โˆต๐‘ก=๐‘ฅ^2+4) Solving ๐‘ฐ๐Ÿ ๐ผ2=โˆซ1โ–’ใ€–3/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅใ€— =3โˆซ1โ–’1/(๐‘ฅ^2+4) ๐‘‘๐‘ฅ =3โˆซ1โ–’1/(๐‘ฅ^2 + 2^2 ) ๐‘‘๐‘ฅ =3 ร— 1/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— =3/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— Therefore F(๐‘ฅ)= ๐ผ1+๐ผ2 F(๐‘ฅ)=3๐‘™๐‘œ๐‘”|๐‘ฅ^2+4|+3/2 tan^(โˆ’1)โกใ€–๐‘ฅ/2ใ€— ( As โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/(๐‘ฅ^2 + ๐‘Ž^2 )=1/๐‘Ž tan^(โˆ’1)โกใ€–๐‘ฅ/๐‘Žใ€— ใ€—) Now, โˆซ_0^2โ–’ใ€–(6๐‘ฅ + 3)/(๐‘ฅ^2 + 4) ๐‘‘๐‘ฅ=๐น(2)โˆ’๐น(0) ใ€— =3๐‘™๐‘œ๐‘”|2^2+4|+3/2 tan^(โˆ’1)โกใ€–2/2โˆ’3๐‘™๐‘œ๐‘”|0+4|โˆ’3/2 tan^(โˆ’1)โก(0/2) ใ€— =3๐‘™๐‘œ๐‘”|4+4|+3/2 tan^(โˆ’1)โกใ€–1โˆ’3๐‘™๐‘œ๐‘”|4|โˆ’3/2 ร— 0ใ€— =3๐‘™๐‘œ๐‘”|8|โˆ’3๐‘™๐‘œ๐‘”|4|+3/2 ๐œ‹/4 =3(๐‘™๐‘œ๐‘”|8|โˆ’๐‘™๐‘œ๐‘”|4|)+3๐œ‹/8 =3๐‘™๐‘œ๐‘”|8/4|+3๐œ‹/8 =๐Ÿ‘ ๐ฅ๐จ๐ โกใ€–๐Ÿ+๐Ÿ‘๐…/๐Ÿ–ใ€— (As log A โ€“ log B = log A/B )

Chapter 7 Class 12 Integrals
Concept wise

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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.