Example 9 (ii) - Chapter 7 Class 12 Integrals
Last updated at December 16, 2024 by Teachoo
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Example 1 (i)
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Chapter 7 Class 12 Integrals
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Example 9 Find the following integrals: (ii) ā«1āšš„/(ć3š„ć^2ā13š„ + 10) ā«1āšš„/(ć3š„ć^2 ā 13š„ + 10) Solving denominator ć3š„ć^2+13š„ā10 =3(š„^2+13/3 š„ ā10/3) =3(š„^2+2. š„Ć 13/6 ā10/3) Adding and subtracting (13/6)^2 =3(š„^2+2. š„Ć 13/6+(13/6)^2ā10/3ā(13/6)^2 ) =3((š„+13/6)^2ā10/3ā(169/36)) =3((š„+13/6)^2ā(10/3 +169/36)) =3((š„+13/6)^2ā((120 +169)/36 )) =3((š„+13/6)^2ā289/36) =3((š„+13/6)^2ā(17/6)^2 ) Hence, our equation becomes ā«1āšš„/(ć3š„ć^2 ā 13š„ + 10) = 1/3 ā«1āšš„/((š„ + 13/6)^2ā (17/6)^2 ) It is of form ā«1āćšš„/(š„^2 ā š^2 )=1/2š ššš|(š„ ā š)/(š„ + š)|+š¶1ć Replacing š„ by (š„+13/6)ššš š šš¦ 17/6, = 1/3 Ć 1/2(17/6) Ćlogā”|(š„ + 13/6 ā 17/6)/(š„+ 13/6 + 17/6)| + C = 1/3 Ć 6/2(17) Ćlogā”|((6š„ + 13 ā 17)/6)/((6š„ +13 + 17)/6)| + C = 1/17 logā”|(6š„ ā 4)/(6š„ + 30)| + C = 1/17 logā”|(2(3š„ ā 2))/(6(š„ + 5))|+ C = 1/17 logā”|( (3š„ ā 2))/(3(š„ + 5))|+ C = 1/17 logā”|( (3š„ ā 2))/((š„ + 5))|ā1/17 logā”3 + C = š/šš šššā”|( (šš ā š))/((š + š))|+ C1
