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Ex 7.9, 19 - Direct Integrate 6x + 3 / x2 + 4 dx from 0 to 2

Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.9, 19 - Chapter 7 Class 12 Integrals - Part 4

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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 )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.