Integration Full Chapter Explained - Integration Class 12 - Everything you need

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

Transcript

Ex 7.4, 23 Integrate (5๐‘ฅ + 3)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) โˆซ1โ–’(5๐‘ฅ + 3)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) ๐‘‘๐‘ฅ =5โˆซ1โ–’(๐‘ฅ + 3/5)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) =5/2 โˆซ1โ–’(2๐‘ฅ + 6/5)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) =5/2 โˆซ1โ–’(2๐‘ฅ + 4 โˆ’ 4 + 6/5)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) =5/2 โˆซ1โ–’(2๐‘ฅ + 4 โˆ’ 14/5)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) Rough (๐‘ฅ^2+4๐‘ฅ+10)^โ€ฒ=2๐‘ฅ+4 =5/2 โˆซ1โ–’(2๐‘ฅ + 4)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) ๐‘‘๐‘ฅ+5/2 โˆซ1โ–’(โˆ’ 14/5)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) ๐‘‘๐‘ฅ =5/2 โˆซ1โ–’(2๐‘ฅ + 4)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) ๐‘‘๐‘ฅโˆ’7โˆซ1โ–’๐‘‘๐‘ฅ/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) ๐‘‘๐‘ฅ Solving ๐‘ฐ๐Ÿ I1=5/2 โˆซ1โ–’( 2๐‘ฅ + 4)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) . ๐‘‘๐‘ฅ Let ๐‘ฅ^2 + 4๐‘ฅ + 10=๐‘ก Diff both sides w.r.t.x 2๐‘ฅ+4+0=๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(2๐‘ฅ + 4) Thus, our equation becomes I1=5/2 โˆซ1โ–’( 2๐‘ฅ + 4)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) . ๐‘‘๐‘ฅ Putting the value of (๐‘ฅ^2+4๐‘ฅ+10)=๐‘ก and ๐‘‘๐‘ฅ=๐‘‘๐‘ก/(2๐‘ฅ + 4) I1=5/2 โˆซ1โ–’(2๐‘ฅ + 4)/โˆš๐‘ก . ๐‘‘๐‘ฅ I1=5/2 โˆซ1โ–’(2๐‘ฅ + 4)/โˆš๐‘ก .๐‘‘๐‘ก/(2๐‘ฅ + 4) I1=5/2 โˆซ1โ–’1/โˆš๐‘ก . ๐‘‘๐‘ก I1=5/2 โˆซ1โ–’1/(๐‘ก)^(1/2) . ๐‘‘๐‘ก I1=5/2 โˆซ1โ–’(๐‘ก)^((โˆ’ 1)/2) . ๐‘‘๐‘ก I1=5/2 ใ€–๐‘ก ใ€—^((โˆ’1)/2 + 1)/((โˆ’1)/2 + 1) +๐ถ1 I1=5/2 (๐‘ก ^(1/2 ))/(1/2) +๐ถ1 I1=5 ๐‘ก ^(1/2 )+๐ถ1 I1=5 โˆš๐‘ก+๐ถ1 I1=5 โˆš(๐‘ฅ^2+4๐‘ฅ+10)+๐ถ1 (Using ๐‘ก=๐‘ฅ^2+4๐‘ฅ+1) Solving ๐‘ฐ๐Ÿ I2=โˆซ1โ–’( 7)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš(๐‘ฅ^2 + 2(2)(๐‘ฅ) + 10) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš(๐‘ฅ^2 + 2(2)(๐‘ฅ) + (2)^2 โˆ’ (2)^2 + 10) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš((๐‘ฅ + 2)^2 โˆ’ (2)^2 + 10) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš((๐‘ฅ + 2)^2 โˆ’ 4 + 10) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš((๐‘ฅ + 2)^2 + 6) . ๐‘‘๐‘ฅ I2=7โˆซ1โ–’1/โˆš((๐‘ฅ + 2)^2 + (โˆš6 )^2 ) . ๐‘‘๐‘ฅ I2=7[๐‘™๐‘œ๐‘”โก|๐‘ฅ+2+โˆš((๐‘ฅ+2)^2 + (โˆš6)^2 )| ]+๐ถ2 I2=7 ๐‘™๐‘œ๐‘”โก|๐‘ฅ+2+โˆš(๐‘ฅ^2+4๐‘ฅ+4+6)|+๐ถ2 I2=7 ๐‘™๐‘œ๐‘”โก|๐‘ฅ+2+โˆš(๐‘ฅ^2+4๐‘ฅ+10)|+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+2) and a by โˆš6 , we get Putting the values of I1 and I2 in (1) โˆซ1โ–’(5๐‘ฅ + 3)/โˆš(๐‘ฅ^2 + 4๐‘ฅ + 10) . ๐‘‘๐‘ฅ = ๐ผ_1โˆ’๐ผ_2 =5 โˆš(๐‘ฅ^2+4๐‘ฅ+10)+๐ถ1โˆ’7 ๐‘™๐‘œ๐‘”โก|๐‘ฅ+2+โˆš(๐‘ฅ^2+4๐‘ฅ+10)|+๐ถ2 =๐Ÿ“ โˆš(๐’™^๐Ÿ+๐Ÿ’๐’™+๐Ÿ๐ŸŽ)โˆ’๐Ÿ• ๐’๐’๐’ˆโก|๐’™+๐Ÿ+โˆš(๐’™^๐Ÿ+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.