Ex 7.8, 13 - Chapter 7 Class 12 Integrals

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Ex 7.8, 13 โˆซ_2^3โ–’๐‘ฅ๐‘‘๐‘ฅ/(๐‘ฅ2+1) Step 1 :- Let F(๐‘ฅ)=โˆซ1โ–’ใ€–๐‘ฅ/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅใ€— Let ๐‘ฅ^2 + 1=๐‘ก Differentiating w.r.t.๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ^2+1)=๐‘‘๐‘ก/๐‘‘๐‘ฅ 2๐‘ฅ=๐‘‘๐‘ก/๐‘‘๐‘ฅ d๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Putting Values of ๐‘ฅ^2+1=๐‘ก & ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ โˆซ1โ–’ใ€–(๐‘ฅ ๐‘‘๐‘ฅ)/(๐‘ฅ^2 + 1)=โˆซ1โ–’ใ€–๐‘ฅ/๐‘ก ๐‘‘๐‘ก/2๐‘ฅใ€—ใ€— =1/2 โˆซ1โ–’๐‘‘๐‘ก/๐‘ก =1/2 ๐‘™๐‘œ๐‘”|๐‘ก| =1/2 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+1| Hence F(๐‘ฅ)=1/2 ๐‘™๐‘œ๐‘”|๐‘ฅ^2+1| Step 2 :- โˆซ_2^3โ–’(๐‘ฅ ๐‘‘๐‘ฅ)/(๐‘ฅ^2+1)=๐น(3)โˆ’๐น(2) =1/2 ๐‘™๐‘œ๐‘”|3^2+1|โˆ’1/2 ๐‘™๐‘œ๐‘”|2^2+1| =1/2 logโกใ€–10โˆ’1/2 ๐‘™๐‘œ๐‘”(5)ใ€— =1/2 (logโกใ€–10โˆ’logโก5 ใ€— ) =1/2 ๐‘™๐‘œ๐‘” 10/5 =๐Ÿ/๐Ÿ ๐ฅ๐จ๐ โก๐Ÿ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo