Examples

Chapter 7 Class 12 Integrals
Serial order wise

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Example 9 Find the following integrals: (ii) β«1βππ₯/(γ3π₯γ^2β13π₯ + 10) β«1βππ₯/(γ3π₯γ^2 β 13π₯ + 10) Solving denominator γ3π₯γ^2+13π₯β10 =3(π₯^2+13/3 π₯ β10/3) =3(π₯^2+2. π₯Γ 13/6 β10/3) Adding and subtracting (13/6)^2 =3(π₯^2+2. π₯Γ 13/6+(13/6)^2β10/3β(13/6)^2 ) =3((π₯+13/6)^2β10/3β(169/36)) =3((π₯+13/6)^2β(10/3 +169/36)) =3((π₯+13/6)^2β((120 +169)/36 )) =3((π₯+13/6)^2β289/36) =3((π₯+13/6)^2β(17/6)^2 ) Hence, our equation becomes β«1βππ₯/(γ3π₯γ^2 β 13π₯ + 10) = 1/3 β«1βππ₯/((π₯ + 13/6)^2β (17/6)^2 ) It is of form β«1βγππ₯/(π₯^2 β π^2 )=1/2π πππ|(π₯ β π)/(π₯ + π)|+πΆ1γ Replacing π₯ by (π₯+13/6)πππ π ππ¦ 17/6, = 1/3 Γ 1/2(17/6) Γlogβ‘|(π₯ + 13/6 β 17/6)/(π₯+ 13/6 + 17/6)| + C = 1/3 Γ 6/2(17) Γlogβ‘|((6π₯ + 13 β 17)/6)/((6π₯ +13 + 17)/6)| + C = 1/17 logβ‘|(6π₯ β 4)/(6π₯ + 30)| + C = 1/17 logβ‘|(2(3π₯ β 2))/(6(π₯ + 5))|+ C = 1/17 logβ‘|( (3π₯ β 2))/(3(π₯ + 5))|+ C = 1/17 logβ‘|( (3π₯ β 2))/((π₯ + 5))|β1/17 logβ‘3 + C = π/ππ πππβ‘|( (ππ β π))/((π + π))|+ C1