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Ex 7.4, 19 - Integrate 6x + 7 / root (x - 5) (x - 4) - Ex 7.4

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.4, 19 (6π‘₯ + 7)/√((π‘₯ βˆ’ 5)(π‘₯ βˆ’ 4) ) Integrating the function 𝑀.π‘Ÿ.𝑑.π‘₯ ∫1β–’(6π‘₯ + 7)/√((π‘₯ βˆ’ 5)(π‘₯ βˆ’ 4) ) . 𝑑π‘₯ =∫1β–’(6π‘₯ + 7)/√(π‘₯ (π‘₯ βˆ’ 4) βˆ’5 (π‘₯ βˆ’ 4) ) . 𝑑π‘₯ =∫1β–’(6π‘₯ + 7)/√(π‘₯^2 βˆ’ 4π‘₯ βˆ’ 5π‘₯ + 20) . 𝑑π‘₯ =∫1β–’(6π‘₯ + 7)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ We can write it as ∫1β–’(6π‘₯ + 7)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯=6∫1β–’(π‘₯ + 7/6)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ =6/2 ∫1β–’(2π‘₯ + 14/6)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ =3∫1β–’(2π‘₯ βˆ’ 9 + 7/3 + 9)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ =3∫1β–’(2π‘₯ βˆ’ 9 + 34/3)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ =3∫1β–’(2π‘₯ βˆ’ 9)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯+3Γ—34/3 ∫1▒𝑑π‘₯/√(π‘₯^2 βˆ’ 9π‘₯ + 20) =3∫1β–’(2π‘₯ βˆ’ 9)/√((π‘₯^2 βˆ’ 9π‘₯ + 20) ) . 𝑑π‘₯+34∫1β–’1/√((π‘₯^2 βˆ’ 9π‘₯ + 20) ) . 𝑑π‘₯ Solving π‘°πŸ I1=3∫1β–’(2π‘₯ βˆ’ 9)/√((π‘₯^2 βˆ’ 9π‘₯ + 20) ) . 𝑑π‘₯ Let π‘₯^2βˆ’9π‘₯+20=𝑑 Diff both sides w.r.t. x 2π‘₯βˆ’9+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(2π‘₯ βˆ’ 9) Thus, our equation becomes I1=3∫1β–’(2π‘₯ βˆ’ 9)/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ Put the values of 𝑑=(π‘₯^2βˆ’9π‘₯+20) and 𝑑π‘₯=𝑑𝑑/(2π‘₯ βˆ’ 9) , we get I1=3∫1β–’(2π‘₯ βˆ’ 9)/βˆšπ‘‘ . 𝑑π‘₯ I1=3∫1β–’(2π‘₯ βˆ’ 9)/βˆšπ‘‘ . 𝑑𝑑/(2π‘₯ βˆ’ 9) I1=3∫1β–’1/βˆšπ‘‘ . 𝑑𝑑 I1=3∫1β–’1/(𝑑)^(1/2) . 𝑑𝑑 I1=3∫1β–’(𝑑)^((βˆ’1)/2) . 𝑑𝑑 I1=3 〖𝑑 γ€—^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) +𝐢1 I1=3 〖𝑑 γ€—^(1/2 )/(1/2) +𝐢1 I1=3 . 2 〖𝑑 γ€—^(1/2 )+𝐢1 I1=6 〖𝑑 γ€—^(1/2 )+𝐢1 I1=6 βˆšπ‘‘+𝐢1 I1=6√(π‘₯^2βˆ’9π‘₯+20)+𝐢1 Solving π‘°πŸ, I2=34∫1β–’1/√(π‘₯^2 βˆ’ 9π‘₯ + 20) . 𝑑π‘₯ I2=34∫1β–’1/√(π‘₯^2 βˆ’2(π‘₯)(9/2) + 20) . 𝑑π‘₯ I2=34∫1β–’1/√(π‘₯^2 βˆ’ 2(π‘₯)(9/2) + (9/2)^2βˆ’ (9/2)^2+ 20) . 𝑑π‘₯ I2=34∫1β–’1/√((π‘₯ βˆ’ 9/2)^2 βˆ’ (9/2)^2+ 20) . 𝑑π‘₯ I2=34∫1β–’1/√((π‘₯ βˆ’ 9/2)^2 βˆ’ 81/4 + 20) . 𝑑π‘₯ I2=34∫1β–’1/√((π‘₯ βˆ’ 9/2)^2 βˆ’ 1/4) . 𝑑π‘₯ I2=34∫1β–’1/√((π‘₯ βˆ’ 9/2)^2 βˆ’ (1/2)^2 ) . 𝑑π‘₯ =34[π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√((π‘₯ βˆ’ 9/2)^2+(1/2)^2 )| ]+𝐢2 =34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2+(9/2)^2βˆ’2(π‘₯)(9/2)βˆ’1/4) |+𝐢2 =34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2+ 81/4βˆ’9π‘₯βˆ’1/4) |+𝐢2 =34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2βˆ’9π‘₯+ 81/4βˆ’1/4) |+𝐢2 =34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2βˆ’9π‘₯+ 80/4) |+𝐢2 =34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2βˆ’9π‘₯+20) |+𝐢2 Now, putting values of I1 and I2 in eq. 1 ∫1β–’(6π‘₯ + 7)/√((π‘₯ βˆ’ 2)(π‘₯ βˆ’ 4) ) . 𝑑π‘₯ =3∫1β–’(2π‘₯ βˆ’ 9)/√((π‘₯^2 βˆ’ 9π‘₯ + 20) ) 𝑑π‘₯+34∫1β–’1/√((π‘₯^2 βˆ’ 9π‘₯ + 20) ) 𝑑π‘₯ =6√(π‘₯^2βˆ’9π‘₯+20)+𝐢1+34 π‘™π‘œπ‘”β‘|π‘₯βˆ’ 9/2+√(π‘₯^2βˆ’9π‘₯+20) |+𝐢2 =πŸ”βˆš(𝒙^πŸβˆ’πŸ—π’™+𝟐𝟎) +πŸ‘πŸ’ π’π’π’ˆβ‘|π’™βˆ’ πŸ—/𝟐+√(𝒙^πŸβˆ’πŸ—π’™+𝟐𝟎) |+π‘ͺ

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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