Ex 7.2, 7 - Integrate x root (x+2) - Teachoo Maths - Ex 7.2

Ex 7.2, 7 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.2, 7 - Chapter 7 Class 12 Integrals - Part 3

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Ex 7.2, 7 Integrate the function: ๐‘ฅโˆš(๐‘ฅ+2) ๐‘ฅโˆš(๐‘ฅ+2) Step 1: Let (๐‘ฅ+2)=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‘ฅโˆš(๐‘ฅ+2)ใ€— .๐‘‘๐‘ฅ Putting the value of ๐‘ฅ+2 & ๐‘‘๐‘ฅ . = โˆซ1โ–’ใ€–๐‘ฅโˆš๐‘กใ€— .๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–๐‘ฅโˆš๐‘กใ€— .๐‘‘๐‘ก = โˆซ1โ–’ใ€–(๐‘กโˆ’2) โˆš๐‘กใ€— .๐‘‘๐‘ก = โˆซ1โ–’ใ€–(๐‘กโˆ’2) ๐‘ก^(1/2) ใ€— .๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก.๐‘ก^(1/2)โˆ’2.๐‘ก^(1/2) ) .๐‘‘๐‘ก = โˆซ1โ–’(๐‘ก^(3/2)โˆ’2.๐‘ก^(1/2) ) .๐‘‘๐‘ก = โˆซ1โ–’๐‘ก^(3/2) .๐‘‘๐‘ก โˆ’ 2โˆซ1โ–’๐‘ก^(1/2) .๐‘‘๐‘ก (Using ๐‘ฅ+2=๐‘ก, ๐‘ฅ=๐‘กโˆ’2) = ๐‘ก^(3/2 + 1)/(3/2 + 1) โˆ’ 2 . ๐‘ก^(1/2 + 1)/(1/2 + 1) + ๐ถ = ๐‘ก^(5/2)/(5/2) โˆ’ 2 . ๐‘ก^(3/2)/(3/2) + ๐ถ = 2/5 ๐‘ก^(5/2) โˆ’ 2 ร— 2/3 ๐‘ก^(3/2) + ๐ถ = 2/5 ๐‘ก^(5/2) โˆ’ 4/3 ๐‘ก^(3/2) + ๐ถ Putting back ๐‘ก=๐‘ฅ+2 = ๐Ÿ/๐Ÿ“ (๐’™+๐Ÿ)^(๐Ÿ“/๐Ÿ) โˆ’ ๐Ÿ’/๐Ÿ‘ (๐’™+๐Ÿ)^(๐Ÿ‘/๐Ÿ) + ๐‘ช (Using โˆซ1โ–’๐‘ฅ^๐‘› . ๐‘‘๐‘ฅ=๐‘ฅ^(๐‘›+1)/(๐‘› +1) )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.