Integration by specific formulaes - Method 10

Chapter 7 Class 12 Integrals
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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