Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

Transcript

Ex 7.7, 8 √(π‘₯2+3π‘₯) ∫1β–’γ€–βˆš(π‘₯^2+3π‘₯) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(π‘₯^2+2(3/2)(π‘₯) ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš(π‘₯^2+2(3/2)(π‘₯)+(3/2)^2βˆ’(3/2)^2 ) 𝑑π‘₯γ€— =∫1β–’γ€–βˆš((π‘₯+3/2)^2βˆ’(3/2)^2 ) 𝑑π‘₯γ€— It is of the form ∫1β–’γ€–βˆš(π‘₯^2βˆ’π‘Ž^2 ) 𝑑π‘₯=π‘₯/2 √(π‘₯^2βˆ’π‘Ž^2 )βˆ’π‘Ž^2/2 π‘™π‘œπ‘”|π‘₯+√(π‘₯^2βˆ’π‘Ž^2 )|+𝐢〗 ∴ Replacing π‘₯ by π‘₯+3/2 and a by 3/2 , we get =(π‘₯ + 3/2)/2 √((π‘₯+3/2)^2βˆ’(3/2)^2 )βˆ’|π‘₯+3/2+√((π‘₯+π‘₯/2)^2βˆ’(3/2)^2 )|+𝐢 =(2π‘₯ + 3)/4 √((π‘₯+3/2)^2βˆ’9/4) βˆ’ 9/8 π‘™π‘œπ‘”|π‘₯+3/2 √((π‘₯+3/2)^2βˆ’9/4)|+𝐢 =(2π‘₯ + 3)/4 √(π‘₯^2+9/4+2(π‘₯)(3/2)βˆ’9/4) βˆ’ 9/8 π‘™π‘œπ‘”|π‘₯+3/2+√(π‘₯^2+9/4+2 (π‘₯)(3/2)βˆ’9/4)|+𝐢 =(2π‘₯ + 3)/4 √(π‘₯^2+2π‘₯(3/2) )βˆ’ 9/8 π‘™π‘œπ‘”|π‘₯+3/2+√(π‘₯^2+2π‘₯(3/2) )|+𝐢 =(πŸπ’™ + πŸ‘)/πŸ’ √(𝒙^𝟐+πŸ‘π’™)βˆ’ πŸ—/πŸ– π’π’π’ˆ|𝒙+πŸ‘/𝟐+√(𝒙^𝟐+πŸ‘π’™)|+π‘ͺ

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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.