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 7.4, 23 Integrate (5π‘₯ + 3)/√(π‘₯^2 + 4π‘₯ + 10) ∫1β–’(5π‘₯ + 3)/√(π‘₯^2 + 4π‘₯ + 10) 𝑑π‘₯ =5∫1β–’(π‘₯ + 3/5)/√(π‘₯^2 + 4π‘₯ + 10) =5/2 ∫1β–’(2π‘₯ + 6/5)/√(π‘₯^2 + 4π‘₯ + 10) =5/2 ∫1β–’(2π‘₯ + 4 βˆ’ 4 + 6/5)/√(π‘₯^2 + 4π‘₯ + 10) =5/2 ∫1β–’(2π‘₯ + 4 βˆ’ 14/5)/√(π‘₯^2 + 4π‘₯ + 10) Rough (π‘₯^2+4π‘₯+10)^β€²=2π‘₯+4 =5/2 ∫1β–’(2π‘₯ + 4)/√(π‘₯^2 + 4π‘₯ + 10) 𝑑π‘₯+5/2 ∫1β–’(βˆ’ 14/5)/√(π‘₯^2 + 4π‘₯ + 10) 𝑑π‘₯ =5/2 ∫1β–’(2π‘₯ + 4)/√(π‘₯^2 + 4π‘₯ + 10) 𝑑π‘₯βˆ’7∫1▒𝑑π‘₯/√(π‘₯^2 + 4π‘₯ + 10) 𝑑π‘₯ Solving π‘°πŸ I1=5/2 ∫1β–’( 2π‘₯ + 4)/√(π‘₯^2 + 4π‘₯ + 10) . 𝑑π‘₯ Let π‘₯^2 + 4π‘₯ + 10=𝑑 Diff both sides w.r.t.x 2π‘₯+4+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(2π‘₯ + 4) Thus, our equation becomes I1=5/2 ∫1β–’( 2π‘₯ + 4)/√(π‘₯^2 + 4π‘₯ + 10) . 𝑑π‘₯ Putting the value of (π‘₯^2+4π‘₯+10)=𝑑 and 𝑑π‘₯=𝑑𝑑/(2π‘₯ + 4) I1=5/2 ∫1β–’(2π‘₯ + 4)/βˆšπ‘‘ . 𝑑π‘₯ I1=5/2 ∫1β–’(2π‘₯ + 4)/βˆšπ‘‘ .𝑑𝑑/(2π‘₯ + 4) I1=5/2 ∫1β–’1/βˆšπ‘‘ . 𝑑𝑑 I1=5/2 ∫1β–’1/(𝑑)^(1/2) . 𝑑𝑑 I1=5/2 ∫1β–’(𝑑)^((βˆ’ 1)/2) . 𝑑𝑑 I1=5/2 〖𝑑 γ€—^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) +𝐢1 I1=5/2 (𝑑 ^(1/2 ))/(1/2) +𝐢1 I1=5 𝑑 ^(1/2 )+𝐢1 I1=5 βˆšπ‘‘+𝐢1 I1=5 √(π‘₯^2+4π‘₯+10)+𝐢1 (Using 𝑑=π‘₯^2+4π‘₯+1) Solving π‘°πŸ I2=∫1β–’( 7)/√(π‘₯^2 + 4π‘₯ + 10) . 𝑑π‘₯ I2=7∫1β–’1/√(π‘₯^2 + 2(2)(π‘₯) + 10) . 𝑑π‘₯ I2=7∫1β–’1/√(π‘₯^2 + 2(2)(π‘₯) + (2)^2 βˆ’ (2)^2 + 10) . 𝑑π‘₯ I2=7∫1β–’1/√((π‘₯ + 2)^2 βˆ’ (2)^2 + 10) . 𝑑π‘₯ I2=7∫1β–’1/√((π‘₯ + 2)^2 βˆ’ 4 + 10) . 𝑑π‘₯ I2=7∫1β–’1/√((π‘₯ + 2)^2 + 6) . 𝑑π‘₯ I2=7∫1β–’1/√((π‘₯ + 2)^2 + (√6 )^2 ) . 𝑑π‘₯ I2=7[π‘™π‘œπ‘”β‘|π‘₯+2+√((π‘₯+2)^2 + (√6)^2 )| ]+𝐢2 I2=7 π‘™π‘œπ‘”β‘|π‘₯+2+√(π‘₯^2+4π‘₯+4+6)|+𝐢2 I2=7 π‘™π‘œπ‘”β‘|π‘₯+2+√(π‘₯^2+4π‘₯+10)|+𝐢2 It is of form ∫1▒𝑑π‘₯/√(π‘₯^2 + π‘Ž^2 ) =π‘™π‘œπ‘”β‘|π‘₯+√(π‘₯^2 + π‘Ž^2 )|+𝐢2 ∴ Replacing x by (π‘₯+2) and a by √6 , we get Putting the values of I1 and I2 in (1) ∫1β–’(5π‘₯ + 3)/√(π‘₯^2 + 4π‘₯ + 10) . 𝑑π‘₯ = 𝐼_1βˆ’πΌ_2 =5 √(π‘₯^2+4π‘₯+10)+𝐢1βˆ’7 π‘™π‘œπ‘”β‘|π‘₯+2+√(π‘₯^2+4π‘₯+10)|+𝐢2 =πŸ“ √(𝒙^𝟐+πŸ’π’™+𝟏𝟎)βˆ’πŸ• π’π’π’ˆβ‘|𝒙+𝟐+√(𝒙^𝟐+πŸ’π’™+𝟏𝟎)|+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.