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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.4, 19 Integrate (6๐‘ฅ + 7)/โˆš((๐‘ฅ โˆ’ 5)(๐‘ฅ โˆ’ 4) ) โˆซ1โ–’(6๐‘ฅ + 7)/โˆš((๐‘ฅ โˆ’ 5)(๐‘ฅ โˆ’ 4) ) . ๐‘‘๐‘ฅ =โˆซ1โ–’(6๐‘ฅ + 7)/โˆš(๐‘ฅ (๐‘ฅ โˆ’ 4) โˆ’5 (๐‘ฅ โˆ’ 4) ) . ๐‘‘๐‘ฅ =โˆซ1โ–’(6๐‘ฅ + 7)/โˆš(๐‘ฅ^2 โˆ’ 4๐‘ฅ โˆ’ 5๐‘ฅ + 20) . ๐‘‘๐‘ฅ =โˆซ1โ–’(6๐‘ฅ + 7)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ =6โˆซ1โ–’(๐‘ฅ + 7/6)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ Rough (๐‘ฅ^2โˆ’9๐‘ฅ+20)โ€ฒ=2๐‘ฅโˆ’9 =6/2 โˆซ1โ–’(2๐‘ฅ + 14/6)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ =3โˆซ1โ–’(2๐‘ฅ โˆ’ 9 + 7/3 + 9)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ =3โˆซ1โ–’(2๐‘ฅ โˆ’ 9 + 34/3)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ =3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) ๐‘‘๐‘ฅ+3ร—34/3 โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) =3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš((๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) ) ๐‘‘๐‘ฅ+34โˆซ1โ–’๐‘‘๐‘ฅ/โˆš((๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) ) โ€ฆ(1) Solving ๐‘ฐ๐Ÿ I1=3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš((๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) ) . ๐‘‘๐‘ฅ Let ๐‘ฅ^2โˆ’9๐‘ฅ+20=๐‘ก Diff both sides w.r.t. x 2๐‘ฅโˆ’9+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(2๐‘ฅ โˆ’ 9) Thus, our equation becomes I1=3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ Putting values of ๐‘ก=(๐‘ฅ^2โˆ’9๐‘ฅ+20) and ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(2๐‘ฅ โˆ’ 9) , we get I1=3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš๐‘ก . ๐‘‘๐‘ฅ I1=3โˆซ1โ–’(2๐‘ฅ โˆ’ 9)/โˆš๐‘ก . ๐‘‘๐‘ก/(2๐‘ฅ โˆ’ 9) I1=3โˆซ1โ–’1/โˆš๐‘ก . ๐‘‘๐‘ก I1=3โˆซ1โ–’1/(๐‘ก)^(1/2) . ๐‘‘๐‘ก I1=3โˆซ1โ–’(๐‘ก)^((โˆ’1)/2) . ๐‘‘๐‘ก I1=3 ใ€–๐‘ก ใ€—^((โˆ’1)/2 + 1)/((โˆ’1)/2 + 1) +๐ถ1 I1=3 ใ€–๐‘ก ใ€—^(1/2 )/(1/2) +๐ถ1 I1=3 . 2 ใ€–๐‘ก ใ€—^(1/2 )+๐ถ1 I1=6 ใ€–๐‘ก ใ€—^(1/2 )+๐ถ1 I1=6 โˆš๐‘ก+๐ถ1 I1=6โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+20)+๐ถ1 Solving ๐‘ฐ๐Ÿ I2=34โˆซ1โ–’1/โˆš(๐‘ฅ^2 โˆ’ 9๐‘ฅ + 20) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš(๐‘ฅ^2 โˆ’2(๐‘ฅ)(9/2) + 20) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš(๐‘ฅ^2 โˆ’ 2(๐‘ฅ)(9/2) + (9/2)^2โˆ’ (9/2)^2+ 20) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš((๐‘ฅ โˆ’ 9/2)^2 โˆ’ (9/2)^2+ 20) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš((๐‘ฅ โˆ’ 9/2)^2 โˆ’ 81/4 + 20) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš((๐‘ฅ โˆ’ 9/2)^2 โˆ’ 1/4) . ๐‘‘๐‘ฅ I2=34โˆซ1โ–’1/โˆš((๐‘ฅ โˆ’ 9/2)^2 โˆ’ (1/2)^2 ) . ๐‘‘๐‘ฅ It is of form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 โˆ’ ๐‘Ž^2 ) =๐‘™๐‘œ๐‘”โก|๐‘ฅ+โˆš(๐‘ฅ^2โˆ’๐‘Ž^2 )|+๐ถ โˆด Replacing x by (๐‘ฅ โˆ’ 9/2) and a by 1/2 , we get =34[๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš((๐‘ฅ โˆ’ 9/2)^2+(1/2)^2 )| ]+๐ถ2 =34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2+(9/2)^2โˆ’2(๐‘ฅ)(9/2)โˆ’1/4) |+๐ถ2 =34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2+ 81/4โˆ’9๐‘ฅโˆ’1/4) |+๐ถ2 =34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+ 81/4โˆ’1/4) |+๐ถ2 =34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+ 80/4) |+๐ถ2 =34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+20) |+๐ถ2 Now, putting values of I1 and I2 in eq. 1 โˆซ1โ–’(6๐‘ฅ + 7)/โˆš((๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 4) ) . ๐‘‘๐‘ฅ = ๐ผ_1+๐ผ_2 =6โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+20)+๐ถ1+34 ๐‘™๐‘œ๐‘”โก|๐‘ฅโˆ’ 9/2+โˆš(๐‘ฅ^2โˆ’9๐‘ฅ+20) |+๐ถ2 =๐Ÿ”โˆš(๐’™^๐Ÿโˆ’๐Ÿ—๐’™+๐Ÿ๐ŸŽ) +๐Ÿ‘๐Ÿ’ ๐’๐’๐’ˆโก|๐’™โˆ’ ๐Ÿ—/๐Ÿ+โˆš(๐’™^๐Ÿโˆ’๐Ÿ—๐’™+๐Ÿ๐ŸŽ) |+๐‘ช

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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.