Ex 7.4, 23 - Integrate 5x + 3 / root x^2 + 4x + 10 - Teachoo

Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 7


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.