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Ex 7.2, 11 - Integrate x / root (x+4) - Class 12 NCERT - Ex 7.2

Ex 7.2, 11 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 11 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.2, 11 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.2, 11 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.2, 11 - Chapter 7 Class 12 Integrals - Part 6


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Ex 7.2, 11 (Method 1) Integrate the function: ๐‘ฅ/โˆš(๐‘ฅ + 4) , ๐‘ฅ>0 Step 1: Simplify the given function ๐‘ฅ/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4 โˆ’ 4)/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4)/โˆš(๐‘ฅ + 4) โˆ’ 4/โˆš(๐‘ฅ + 4) = (๐‘ฅ + 4)^(1/2 โˆ’ 1) โˆ’ 4(๐‘ฅ + 4)^(1/2) = (๐‘ฅ + 4)^(1/2) โˆ’ 4(๐‘ฅ + 4)^(โˆ’ 1/2) (Adding and Subtracting 4) Step 2: Integrating the function โˆซ1โ–’ใ€–" " ๐‘ฅ/โˆš(๐‘ฅ + 4)ใ€— . ๐‘‘๐‘ฅ = โˆซ1โ–’((๐‘ฅ + 4)^(1/2) " โˆ’ " ใ€–4 (๐‘ฅ + 4)ใ€—^(โˆ’ 1/2) ) . ๐‘‘๐‘ฅ = โˆซ1โ–’(๐‘ฅ + 4)^(1/2) . ๐‘‘๐‘ฅ โˆ’ 4โˆซ1โ–’(๐‘ฅ + 4)^(โˆ’ 1/2) . ๐‘‘๐‘ฅ = (๐‘ฅ + 4)^(1/2 + 1)/(1/2 + 1) โˆ’ (4 (๐‘ฅ + 4)^(โˆ’ 1/2 + 1))/(โˆ’ 1/2 + 1) + C = (๐‘ฅ + 4)^(3/2)/(3/2 ) โˆ’ (4 (๐‘ฅ + 4)^(1/2))/( 1/2) + C = 2/3 (๐‘ฅ+4)^(3/2) โˆ’ 4.2 (๐‘ฅ+4)^(1/2) + C = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ + 4)/3 โˆ’4) + ๐ถ = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ + 4 โˆ’12)/3) + ๐ถ = ใ€–2(๐‘ฅ+4)ใ€—^(1/2) ((๐‘ฅ โˆ’ 8))/3 + ๐ถ = ๐Ÿ/๐Ÿ‘ โˆš(๐’™ + ๐Ÿ’) (๐’™โˆ’๐Ÿ–) + ๐‘ช (Taking ใ€–2(๐‘ฅ+4)ใ€—^(1/2) as common) Ex 7.2, 11 (Method 2) Integrate the function: ๐‘ฅ/โˆš(๐‘ฅ + 4) , ๐‘ฅ>0 Step 1: Let ๐‘ฅ+4=๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ 1+0= ๐‘‘๐‘ก/๐‘‘๐‘ฅ 1= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=๐‘‘๐‘ก Step 2: Integrating the function โˆซ1โ–’ใ€–" " ๐‘ฅ/โˆš(๐‘ฅ + 4)ใ€— . ๐‘‘๐‘ฅ Putting ๐‘ก=๐‘ฅ+4 & ๐‘‘๐‘ฅ=๐‘‘๐‘ก = โˆซ1โ–’๐‘ฅ/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก โˆ’ 4)/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก โˆ’ 4)/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก/โˆš๐‘ก โˆ’4/โˆš๐‘ก) ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก/โˆš๐‘ก . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’4/โˆš๐‘ก . ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก^(1 โˆ’ 1/2) . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’ใ€–4 . ๐‘ก^(โˆ’ 1/2) ใ€—. ๐‘‘๐‘ก = โˆซ1โ–’๐‘ก^(1/2) . ๐‘‘๐‘ก โˆ’ โˆซ1โ–’ใ€–4 . ๐‘ก^(โˆ’ 1/2) ใ€—. ๐‘‘๐‘ก (As ๐‘ฅ + 4=๐‘ก โ‡’๐‘ฅ=๐‘กโˆ’4) = ๐‘ก^(1/2 + 1)/(1/2 + 1) โˆ’ 4 ๐‘ก^(โˆ’ 1/2 + 1)/(โˆ’ 1/2 + 1) +๐ถ = (๐‘ก^(3/2) )/(3/2) โˆ’ 4 ๐‘ก^(1/2)/(1/2) +๐ถ = 2/3 ๐‘ก^(3/2) โˆ’ 4 . 2๐‘ก^(1/2) +๐ถ Taking 2/3 . ๐‘ก^(1/2) as common , we get = 2/3 . ๐‘ก^(1/2) (๐‘กโˆ’4 . 3)+๐ถ = 2/3 . ๐‘ก^(1/2) (๐‘กโˆ’12)+๐ถ Putting the value of ๐‘ก=๐‘ฅ+4 = 2/3 . (๐‘ฅ+4)^(1/2) (๐‘ฅ+4โˆ’12)+๐ถ = ๐Ÿ/๐Ÿ‘ โˆš(๐’™+๐Ÿ’) (๐’™โˆ’๐Ÿ–)+๐‘ช

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.