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Ex 7.5, 7 - Integrate x / (x2 + 1) (x - 1) - Class 12 NCERT - Integration by partial fraction - Type 5

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 7 ๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) Let I=โˆซ1โ–’๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) ๐‘‘๐‘ฅ We can write integrand as ๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) =(๐ด๐‘ฅ + ๐ต)/((๐‘ฅ^2 + 1) ) + ๐ถ/((๐‘ฅ โˆ’ 1) ) ๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) =((๐ด๐‘ฅ + ๐ต)(๐‘ฅ โˆ’ 1) + ๐ถ(๐‘ฅ^2 + 1))/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) By Cancelling denominator ๐‘ฅ=(๐ด๐‘ฅ+๐ต)(๐‘ฅโˆ’1)+๐ถ(๐‘ฅ^2+1) Putting x = 1, in (1) ๐‘ฅ=(๐ด๐‘ฅ+๐ต)(๐‘ฅโˆ’1)+๐ถ(๐‘ฅ^2+1) 1=(๐ดร—1+๐ต)(1โˆ’1)+๐ถ(1^2+1) 1=(๐ด+๐ต)ร—0+๐ถ(1+1) 1=0+2๐ถ ๐ถ=1/2 Putting x = 0 , in (1) ๐‘ฅ=(๐ด๐‘ฅ+๐ต)(๐‘ฅโˆ’1)+๐ถ(๐‘ฅ^2+1) 0=(๐ดร—0+๐ต)(0โˆ’1)+๐ถ(0+1) 0 =๐ต(โˆ’1)+๐ถ โ‡’๐ต=๐ถ=1/2 Putting x = โˆ’1 โˆ’1=(๐ด(โˆ’1)+๐ต)(โˆ’1โˆ’1)+๐ถ((1)^2+1) โˆ’1=(โˆ’๐ด+๐ต)(โˆ’2)+๐ถ(1+1) โˆ’1=2๐ดโˆ’2๐ต+2๐ถ โˆ’1=2๐ดโˆ’2ร—1/2+2ร—1/2 โˆ’1=2๐ดโˆ’1+1 โˆ’1=2๐ด ๐ด=(โˆ’1)/2 Hence we can write ๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) =((โˆ’ 1/2 ๐‘ฅ + 1/2))/((๐‘ฅ^2 + 1) ) + (1/2)/((๐‘ฅ โˆ’ 1) ) =(โˆ’1 . ๐‘ฅ)/(2 (๐‘ฅ^2 + 1) ) + 1/(2 (๐‘ฅ^2 + 1) )+1/2(๐‘ฅ โˆ’ 1) Integrating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ I=โˆซ1โ–’๐‘ฅ/(๐‘ฅ^2 + 1)(๐‘ฅ โˆ’ 1) ๐‘‘๐‘ฅ =โˆ’1/2 โˆซ1โ–’๐‘ฅ/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅ+1/2 โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ^2 + 1) + 1/2 โˆซ1โ–’๐‘‘๐‘ฅ/(๐‘ฅ โˆ’ 1) ๐‘‘๐‘ฅ Solving ๐ˆ๐Ÿ I1=โˆ’1/2 โˆซ1โ–’๐‘ฅ/((๐‘ฅ^2 + 1) ) ๐‘‘๐‘ฅ Put ๐‘ฅ^2+1=๐‘ก Different both sides w.r.t.๐‘ฅ 2๐‘ฅ+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/2๐‘ฅ Hence we can write โˆ’1/2 โˆซ1โ–’๐‘ฅ/((๐‘ฅ^2 + 1) ) ๐‘‘๐‘ฅ=โˆ’1/2 โˆซ1โ–’๐‘ฅ/๐‘ก ๐‘‘๐‘ก/2๐‘ฅ =โˆ’1/(2 ร— 2) โˆซ1โ–’๐‘‘๐‘ก/๐‘ก =โˆ’1/4 logโก|๐‘ก|+๐ถ1 =โˆ’1/4 logโก|๐‘ฅ^2+1|+๐ถ1 Solving ๐ˆ๐Ÿ I2=1/2 โˆซ1โ–’1/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅ =1/2 tan^(โˆ’1)โก๐‘ฅ+๐ถ2 Solving ๐ˆ๐Ÿ‘ I3=1/2 โˆซ1โ–’1/(๐‘ฅ โˆ’ 1) ๐‘‘๐‘ฅ =1/2 logโก|๐‘ฅโˆ’1|+๐ถ3 Hence I=I1+I2+I3 =โˆ’1/4 logโก|๐‘ฅ^2+1|+๐ถ1+1/2 tan^(โˆ’1)โก๐‘ฅ+๐ถ2+1/2 logโก|๐‘ฅโˆ’1|+๐ถ3 =โˆ’1/4 logโก|๐‘ฅ^2+1|+1/2 tan^(โˆ’1)โก๐‘ฅ+1/2 logโก|๐‘ฅโˆ’1|+๐ถ =โˆ’1/4 logโก|๐‘ฅ^2+1|+1/2 tan^(โˆ’1)โก๐‘ฅ+1/2 logโก|๐‘ฅโˆ’1|+๐ถ =๐Ÿ/๐Ÿ ๐’๐’๐’ˆโก|๐’™โˆ’๐Ÿ|โˆ’๐Ÿ/๐Ÿ’ ใ€–๐’๐’๐’ˆ ใ€—โก|๐’™^๐Ÿ+๐Ÿ|+๐Ÿ/๐Ÿ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก๐’™+ C

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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