Slide7.JPG

Slide8.JPG
Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG

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

Transcript

Ex 7.4, 21 Integrate the function (๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) โˆซ1โ–’(๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(2๐‘ฅ + 4)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(2๐‘ฅ + 2 + 4 โˆ’ 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ+2/2 โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ+โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) ๐‘‘๐‘ฅ Rough (๐‘ฅ^2+2๐‘ฅ+3)^โ€ฒ=2๐‘ฅ+2 Solving ๐‘ฐ๐Ÿ I1=1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) . ๐‘‘๐‘ฅ Let ๐‘ฅ^2 + 2๐‘ฅ + 3=๐‘ก Diff both sides w.r.t.x 2๐‘ฅ+2+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(2๐‘ฅ + 2) Now, our equation becomes I1=1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) . ๐‘‘๐‘ฅ Putting the value of (4๐‘ฅโˆ’๐‘ฅ^2 ) and ๐‘‘๐‘ฅ I1=1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš๐‘ก . ๐‘‘๐‘ฅ I1=1/2 โˆซ1โ–’(2๐‘ฅ + 2)/โˆš๐‘ก . ๐‘‘๐‘ก/(2๐‘ฅ + 2) I1=1/2 โˆซ1โ–’1/โˆš๐‘ก . ๐‘‘๐‘ก I1=1/2 โˆซ1โ–’1/(๐‘ก)^(1/2) . ๐‘‘๐‘ก I1=1/2 โˆซ1โ–’(๐‘ก)^((โˆ’ 1)/2) . ๐‘‘๐‘ก I1=1/2 ใ€–๐‘ก ใ€—^((โˆ’1)/2 + 1)/((โˆ’1)/2 + 1) +๐ถ1 I1= ๐‘ก ^(1/2 )+๐ถ1 I1= โˆš๐‘ก+๐ถ1 I1=โˆš(๐‘ฅ^2+2๐‘ฅ+3)+๐ถ Solving ๐‘ฐ๐Ÿ I2=โˆซ1โ–’1/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3) . ๐‘‘๐‘ฅ I2=โˆซ1โ–’1/โˆš(๐‘ฅ^2 + 2(๐‘ฅ)(1) + 3) . ๐‘‘๐‘ฅ I2=โˆซ1โ–’1/โˆš(๐‘ฅ^2 + 2(๐‘ฅ)(1) โˆ’ (1)^2 + (1)^2 + 3) . ๐‘‘๐‘ฅ (Using ๐‘ก=๐‘ฅ^2+2๐‘ฅ+3) I2=โˆซ1โ–’1/โˆš((๐‘ฅ + 1)^2 โˆ’ (1)^2 + 3) . ๐‘‘๐‘ฅ I2=โˆซ1โ–’1/โˆš((๐‘ฅ + 1)^2 โˆ’ 1 + 3) . ๐‘‘๐‘ฅ I2=โˆซ1โ–’1/โˆš((๐‘ฅ + 1)^2 + 2) . ๐‘‘๐‘ฅ I2=โˆซ1โ–’1/โˆš((๐‘ฅ + 1)^2 +(โˆš2 )^2 ) . ๐‘‘๐‘ฅ I2=๐‘™๐‘œ๐‘”โก|๐‘ฅ+1+โˆš((๐‘ฅ + 1)^2+(โˆš2 )^2 )|+๐ถ2 It is of form โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + ๐‘Ž^2 ) =๐‘™๐‘œ๐‘”โก|๐‘ฅ+โˆš(๐‘ฅ^2 + ๐‘Ž^2 )|+๐ถ2 โˆด Replacing x by (๐‘ฅ + 1) and a by โˆš2 , we get I2=๐‘™๐‘œ๐‘”โก|๐‘ฅ+1+โˆš(๐‘ฅ^2+2๐‘ฅ+1+2)|+๐ถ2 I2=๐‘™๐‘œ๐‘”โก|๐‘ฅ+1+โˆš(๐‘ฅ^2+2๐‘ฅ+3)|+๐ถ2 Putting the values of I1 and I2 in (1) โˆซ1โ–’ใ€–(๐‘ฅ + 2)/โˆš(๐‘ฅ^2 + 2๐‘ฅ + 3).ใ€— . ๐‘‘๐‘ฅ = ๐ผ_1+๐ผ_2 =โˆš(๐‘ฅ^2+2๐‘ฅ+3)+๐ถ1+๐‘™๐‘œ๐‘”โก|๐‘ฅ+1+โˆš(๐‘ฅ^2+2๐‘ฅ+3)|+๐ถ2 =โˆš(๐’™^๐Ÿ+๐Ÿ๐’™+๐Ÿ‘)+๐’๐’๐’ˆโก|๐’™+๐Ÿ+โˆš(๐’™^๐Ÿ+๐Ÿ๐’™+๐Ÿ‘)|+๐‘ช

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.