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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Example 25 (Supplementary NCERT) โˆซ1โ–’๐‘ฅ โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ โˆซ1โ–’๐‘ฅ โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ We can write it as:- x = A [๐‘‘/๐‘‘๐‘ฅ (1+๐‘ฅโˆ’๐‘ฅ^2 )]+ B x = A [0+1โˆ’2๐‘ฅ]+ B x = A [1โˆ’2๐‘ฅ]+ B x = "A"โˆ’2"A" ๐‘ฅ+ B x = (โˆ’2A) x ๐‘ฅ/๐‘ฅ = โˆ’2A 1 = โˆ’2A A = (โˆ’1)/2 0 = A + B B = โˆ’A B = โˆ’((โˆ’1)/2) B = 1/2 Thus, we can write x = A [1โˆ’2๐‘ฅ] + B x = ((โˆ’1)/2)[1โˆ’2๐‘ฅ]Integrating โˆซ1โ–’ใ€–๐‘ฅโˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[(โˆ’1/2)[1โˆ’2๐‘ฅ]+1/2] ใ€— โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[(โˆ’1/2)[1 โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )+1/2 โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )] ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–(โˆ’1/2)[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’ใ€–1/2 โˆš(1โˆ’๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = โˆ’1/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ+ใ€— 1/2 โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving ๐‘ฐ_๐Ÿ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Let 1 + ๐‘ฅ โˆ’ ๐‘ฅ^2 = t Diff. both sides w.r.t.x 0 + 1 โˆ’2x = ๐‘‘๐‘ก/๐‘‘๐‘ฅ (1 โˆ’ 2x) dx = dt dx = ๐‘‘๐‘ก/(1 โˆ’ 2๐‘ฅ) Thus, our equation bceoms I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ Putting the value if (1+๐‘ฅโˆ’๐‘ฅ^2) and dx, we get I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ฅ I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš๐‘กใ€—. ๐‘‘๐‘ก/[1 โˆ’ 2๐‘ฅ] ๐ผ_1 = (โˆ’1)/2 โˆซ1โ–’โˆš๐‘ก. ๐‘‘๐‘ก I_1 = (โˆ’1)/2 โˆซ1โ–’ใ€–(๐‘ก)ใ€—^(1/2) ๐‘‘๐‘ก I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^(1/2 + 1)/((1/2 + 1) )+ C_1 I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^((1 + 2)/2 )/(((1 + 2)/2) )+ C_1 I_1 = (โˆ’1)/2 ใ€–๐‘ก ใ€—^(3/2 )/((3/2) )+ C_1 I_1 = (โˆ’1)/3 ใ€–๐‘ก ใ€—^(3/2 )+ C_1 I_1 = (โˆ’1)/3 ใ€–(1โˆ’๐‘ฅ โˆ’๐‘ฅ^2 ) ใ€—^(3/2 )+ C_1 Solving ๐‘ฐ_๐Ÿ I_2 = 1/2 โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(โˆ’(๐‘ฅ^2โˆ’๐‘ฅโˆ’1)) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(โˆ’[๐‘ฅ^2โˆ’2(๐‘ฅ)(1/2)โˆ’1] ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[๐‘ฅ^2โˆ’2(๐‘ฅ)(1/2)+(1/2)^2โˆ’(1/2)^2โˆ’1] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[(๐‘ฅโˆ’1/2)^2โˆ’(1/2)^2โˆ’1] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2+(โˆ’1 โˆ’4)/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2 โˆ’5/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(5/4 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 ((๐‘ฅ โˆ’ 1/2)/2 โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 )+(โˆš5/2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 1/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2 โˆ’5/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(5/4 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 ((๐‘ฅ โˆ’ 1/2)/2 โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 )+(โˆš5/2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 1/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2 โˆ’5/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(5/4 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 ((๐‘ฅ โˆ’ 1/2)/2 โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 )+(โˆš5/2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 1/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(โˆ’[ใ€– (๐‘ฅโˆ’1/2)ใ€—^2 โˆ’5/4] ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš(5/4 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’ใ€–โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 ) ใ€— ๐‘‘๐‘ฅ I_2 = 1/2 ((๐‘ฅ โˆ’ 1/2)/2 โˆš((โˆš5/4)^2 ใ€– โˆ’(๐‘ฅโˆ’1/2)ใ€—^2 )+(โˆš5/2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((๐‘ฅ + 1/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 (((2๐‘ฅ โˆ’1)/2)/2 โˆš(5/4โˆ’[๐‘ฅ^2 +1/4โˆ’2๐‘ฅ(1/2)] ) +(5/4)/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) (((2๐‘ฅ โˆ’ 1)/2)/(โˆš5/2))+ C_2 ) I_2 = 1/2 ((2๐‘ฅ โˆ’1)/2.2 โˆš(5/4โˆ’[๐‘ฅ^2 +1/4 โˆ’๐‘ฅ] ) +5/(4 . 2) ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+ C_2 ) I_2 " = " (2๐‘ฅ โˆ’1)/8 โˆš(1+๐‘ฅ+๐‘ฅ^2 ) +5/16 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+C_3 Putting the value of I_1 and I_2 in eq. (1) , we get โˆซ1โ–’๐‘ฅ โˆš(1+ ๐‘ฅโˆ’๐‘ฅ^2 ) d๐‘ฅ = (โˆ’1)/2 โˆซ1โ–’ใ€–[1โˆ’2๐‘ฅ] โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ+โˆซ1โ–’โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = (โˆ’1)/3 ใ€–(1+๐‘ฅโˆ’๐‘ฅ^2)ใ€—^(3/2) + C_1 + ((2๐‘ฅ โˆ’1))/8 โˆš(1+๐‘ฅโˆ’๐‘ฅ^2 )+5/16 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1) ((2๐‘ฅ โˆ’ 1)/โˆš5)+ C_3 = (โˆ’๐Ÿ)/๐Ÿ‘ ใ€–(๐Ÿ+๐’™โˆ’๐’™^๐Ÿ)ใ€—^(๐Ÿ‘/๐Ÿ) +๐Ÿ/๐Ÿ– ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) (๐Ÿ๐’™ โˆ’๐Ÿ)(๐Ÿ+๐’™โˆ’๐’™^๐Ÿ)+ ๐Ÿ“/๐Ÿ๐Ÿ” ใ€–๐’”๐’Š๐’ใ€—^(โˆ’๐Ÿ) ((๐Ÿ๐’™ โˆ’ ๐Ÿ)/โˆš๐Ÿ“)+ ๐‚

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
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