Example 9 (ii) - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
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Chapter 7 Class 12 Integrals
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Transcript
Example 9 Find the following integrals: (ii) ∫1▒𝑑𝑥/(〖3𝑥〗^2−13𝑥 + 10) ∫1▒𝑑𝑥/(〖3𝑥〗^2 − 13𝑥 + 10) Solving denominator 〖3𝑥〗^2+13𝑥−10 =3(𝑥^2+13/3 𝑥 −10/3) =3(𝑥^2+2. 𝑥× 13/6 −10/3) Adding and subtracting (13/6)^2 =3(𝑥^2+2. 𝑥× 13/6+(13/6)^2−10/3−(13/6)^2 ) =3((𝑥+13/6)^2−10/3−(169/36)) =3((𝑥+13/6)^2−(10/3 +169/36)) =3((𝑥+13/6)^2−((120 +169)/36 )) =3((𝑥+13/6)^2−289/36) =3((𝑥+13/6)^2−(17/6)^2 ) Hence, our equation becomes ∫1▒𝑑𝑥/(〖3𝑥〗^2 − 13𝑥 + 10) = 1/3 ∫1▒𝑑𝑥/((𝑥 + 13/6)^2− (17/6)^2 ) It is of form ∫1▒〖𝑑𝑥/(𝑥^2 − 𝑎^2 )=1/2𝑎 𝑙𝑜𝑔|(𝑥 − 𝑎)/(𝑥 + 𝑎)|+𝐶1〗 Replacing 𝑥 by (𝑥+13/6)𝑎𝑛𝑑 𝑎 𝑏𝑦 17/6, = 1/3 × 1/2(17/6) ×log|(𝑥 + 13/6 − 17/6)/(𝑥+ 13/6 + 17/6)| + C = 1/3 × 6/2(17) ×log|((6𝑥 + 13 − 17)/6)/((6𝑥 +13 + 17)/6)| + C = 1/17 log|(6𝑥 − 4)/(6𝑥 + 30)| + C = 1/17 log|(2(3𝑥 − 2))/(6(𝑥 + 5))|+ C = 1/17 log|( (3𝑥 − 2))/(3(𝑥 + 5))|+ C = 1/17 log|( (3𝑥 − 2))/((𝑥 + 5))|−1/17 log3 + C = 𝟏/𝟏𝟕 𝒍𝒐𝒈|( (𝟑𝒙 − 𝟐))/((𝒙 + 𝟓))|+ C1
