Check sibling questions

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

Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.4, 19 - Chapter 7 Class 12 Integrals - Part 8


Transcript

Ex 7.4, 19 Integrate (6𝑥 + 7)/√((𝑥 − 5)(𝑥 − 4) ) ∫1▒(6𝑥 + 7)/√((𝑥 − 5)(𝑥 − 4) ) . 𝑑𝑥 =∫1▒(6𝑥 + 7)/√(𝑥 (𝑥 − 4) −5 (𝑥 − 4) ) . 𝑑𝑥 =∫1▒(6𝑥 + 7)/√(𝑥^2 − 4𝑥 − 5𝑥 + 20) . 𝑑𝑥 =∫1▒(6𝑥 + 7)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 =6∫1▒(𝑥 + 7/6)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 Rough (𝑥^2−9𝑥+20)′=2𝑥−9 =6/2 ∫1▒(2𝑥 + 14/6)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 =3∫1▒(2𝑥 − 9 + 7/3 + 9)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 =3∫1▒(2𝑥 − 9 + 34/3)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 =3∫1▒(2𝑥 − 9)/√(𝑥^2 − 9𝑥 + 20) 𝑑𝑥+3×34/3 ∫1▒𝑑𝑥/√(𝑥^2 − 9𝑥 + 20) =3∫1▒(2𝑥 − 9)/√((𝑥^2 − 9𝑥 + 20) ) 𝑑𝑥+34∫1▒𝑑𝑥/√((𝑥^2 − 9𝑥 + 20) ) …(1) Solving 𝑰𝟏 I1=3∫1▒(2𝑥 − 9)/√((𝑥^2 − 9𝑥 + 20) ) . 𝑑𝑥 Let 𝑥^2−9𝑥+20=𝑡 Diff both sides w.r.t. x 2𝑥−9+0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(2𝑥 − 9) Thus, our equation becomes I1=3∫1▒(2𝑥 − 9)/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 Putting values of 𝑡=(𝑥^2−9𝑥+20) and 𝑑𝑥=𝑑𝑡/(2𝑥 − 9) , we get I1=3∫1▒(2𝑥 − 9)/√𝑡 . 𝑑𝑥 I1=3∫1▒(2𝑥 − 9)/√𝑡 . 𝑑𝑡/(2𝑥 − 9) I1=3∫1▒1/√𝑡 . 𝑑𝑡 I1=3∫1▒1/(𝑡)^(1/2) . 𝑑𝑡 I1=3∫1▒(𝑡)^((−1)/2) . 𝑑𝑡 I1=3 〖𝑡 〗^((−1)/2 + 1)/((−1)/2 + 1) +𝐶1 I1=3 〖𝑡 〗^(1/2 )/(1/2) +𝐶1 I1=3 . 2 〖𝑡 〗^(1/2 )+𝐶1 I1=6 〖𝑡 〗^(1/2 )+𝐶1 I1=6 √𝑡+𝐶1 I1=6√(𝑥^2−9𝑥+20)+𝐶1 Solving 𝑰𝟐 I2=34∫1▒1/√(𝑥^2 − 9𝑥 + 20) . 𝑑𝑥 I2=34∫1▒1/√(𝑥^2 −2(𝑥)(9/2) + 20) . 𝑑𝑥 I2=34∫1▒1/√(𝑥^2 − 2(𝑥)(9/2) + (9/2)^2− (9/2)^2+ 20) . 𝑑𝑥 I2=34∫1▒1/√((𝑥 − 9/2)^2 − (9/2)^2+ 20) . 𝑑𝑥 I2=34∫1▒1/√((𝑥 − 9/2)^2 − 81/4 + 20) . 𝑑𝑥 I2=34∫1▒1/√((𝑥 − 9/2)^2 − 1/4) . 𝑑𝑥 I2=34∫1▒1/√((𝑥 − 9/2)^2 − (1/2)^2 ) . 𝑑𝑥 It is of form ∫1▒𝑑𝑥/√(𝑥^2 − 𝑎^2 ) =𝑙𝑜𝑔⁡|𝑥+√(𝑥^2−𝑎^2 )|+𝐶 ∴ Replacing x by (𝑥 − 9/2) and a by 1/2 , we get =34[𝑙𝑜𝑔⁡|𝑥− 9/2+√((𝑥 − 9/2)^2+(1/2)^2 )| ]+𝐶2 =34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2+(9/2)^2−2(𝑥)(9/2)−1/4) |+𝐶2 =34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2+ 81/4−9𝑥−1/4) |+𝐶2 =34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2−9𝑥+ 81/4−1/4) |+𝐶2 =34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2−9𝑥+ 80/4) |+𝐶2 =34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2−9𝑥+20) |+𝐶2 Now, putting values of I1 and I2 in eq. 1 ∫1▒(6𝑥 + 7)/√((𝑥 − 2)(𝑥 − 4) ) . 𝑑𝑥 = 𝐼_1+𝐼_2 =6√(𝑥^2−9𝑥+20)+𝐶1+34 𝑙𝑜𝑔⁡|𝑥− 9/2+√(𝑥^2−9𝑥+20) |+𝐶2 =𝟔√(𝒙^𝟐−𝟗𝒙+𝟐𝟎) +𝟑𝟒 𝒍𝒐𝒈⁡|𝒙− 𝟗/𝟐+√(𝒙^𝟐−𝟗𝒙+𝟐𝟎) |+𝑪

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.