Chapter 7 Class 12 Integrals
Concept wise

Ex 7.5, 2 - Integrate 1 / x2 - 9 - Chapter 7 Class 12 NCERT - Ex 7.5

Ex 7.5, 2 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 2 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.5, 2 1/(𝑥2− 9) Solving integrand 1/(𝑥2− 9)=1/((𝑥 − 3) (𝑥 + 3) ) We can write it as 1/((𝑥 − 3) (𝑥 + 3) )=𝐴/((𝑥 − 3) ) + 𝐵/((𝑥 + 3) ) 1/((𝑥 − 3) (𝑥 + 3) )=(𝐴(𝑥 + 3) + 𝐵(𝑥 − 3))/((𝑥 − 3) (𝑥 + 3) ) " " Cancelling denominator 1 = 𝐴(𝑥 + 3) + 𝐵(𝑥 − 3) Putting 𝑥 = 3 in (1) 1 = 𝐴(𝑥+3)+𝐵(𝑥−3) 1 = 𝐴(3+3) + 𝐵(3−3) 1 = 𝐴×6+ 𝐵×0 1 = 6𝐴 𝐴 = 1/6 Similarly Putting y=−3 in (1) 1 = 𝐴(−3+3) + 𝐵(−3−3) 1 = 𝐴×0+ 𝐵×(−6) 1 = −6𝐵 𝐵 = (−1)/6 Hence we can write it as 1/((𝑥 − 3) (𝑥 + 3) ) = 1/6(𝑥 − 3) − 1/6(𝑥 + 3) Therefore ∫1▒1/((𝑥 − 3) (𝑥 + 3) ) 𝑑𝑥 = 1/6 ∫1▒1/((𝑥 − 3) ) 𝑑𝑥 − 1/6 ∫1▒1/((𝑥 + 3) ) 𝑑𝑥 = 1/6 〖log 〗⁡|𝑥−3|− 1/6 〖log 〗⁡|𝑥+3|+𝐶 = 1/6 (〖log 〗⁡|𝑥−3|−〖log 〗⁡|𝑥+3| )+𝐶 = 𝟏/𝟔 〖𝒍𝒐𝒈 〗⁡|(𝒙 − 𝟑)/(𝒙 + 𝟑)|+𝑪

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.