CBSE Class 12 Sample Paper for 2024 Boards

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Question 26 - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by

Find : ∫(2x^2+3)/(x^2 (x^2+9) ) dx;x≠0

This question is similar to Ex 7.5, 18 Chapter 7 Class 12

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Transcript

(2𝑥^2 + 3)/(𝑥^2 (𝑥^2 + 9) ) Let t = 𝒙^𝟐 = (2𝑡 + 3)/𝑡(𝑡 + 9) We can write (2𝑡 + 3)/𝑡(𝑡 + 9) = 𝑨/𝒕 + 𝑩/((𝒕 + 𝟗) ) (2𝑡 + 3)/𝑡(𝑡 + 9) = (𝐴(𝑡 + 9) +𝐵𝑡)/(𝑡(𝑡 + 9) ) Cancelling denominator 𝟐𝒕+𝟑 = 𝑨(𝒕+𝟗)+𝑩𝒕 Putting t = −𝟗 in (1) 2(−9)+3 = 𝐴(−9+9)+𝐵(−9) −18+3 = 𝐴×0+𝐵(−9) −15 = 0+𝐵(−9) −15 = −9𝐵 B = 15/9 𝑩 = 𝟓/𝟑 Putting t = 𝟎 in (1) 2(0)+3 = 𝐴(0+9)+𝐵(0) 3 = 9𝐴+0 A = 9/3 A = 𝟏/𝟑 Hence we can write (2𝑡 + 3)/𝑡(𝑡 + 9) = (1/3)/𝑡 + (5/3)/((𝑡 + 9 ) ) (2𝑡 + 3)/𝑡(𝑡 + 9) = 1/3𝑡 + 5/(3(𝑡 + 9)) Putting back t = 𝒙^𝟐 (𝟐𝒙^𝟐+ 𝟑)/(𝒙^𝟐 (𝒙^𝟐 + 𝟗) ) = 𝟏/(𝟑𝒙^𝟐 ) + 𝟓/(𝟑(𝒙^(𝟐 )+ 𝟗)) Therefore, ∫1▒(2𝑥^2 + 3)/(𝑥^2 (𝑥^2 +9)) 𝑑𝑥 = ∫1▒〖1/(3𝑥^2 ) 𝑑𝑥"+ " ∫1▒5/(3(𝑥^2+ 9))〗 𝑑𝑥 = 1/3 ∫1▒〖1/𝑥^2 𝑑𝑥" + " 5/3 ∫1▒1/((𝑥^2+ 9))〗 𝑑𝑥 = 𝟏/𝟑 ∫1▒〖𝟏/𝒙^𝟐 𝒅𝒙" + " 𝟓/𝟑 ∫1▒𝟏/((𝒙^𝟐+ 〖(𝟑)〗^𝟐))〗 𝒅𝒙 = 1/3 × (−𝟏)/𝒙 + 5/3 × 𝟏/𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖 𝒙/𝟑〗 + C = (−𝟏)/𝟑𝒙 + 𝟓/𝟗 〖𝒕𝒂𝒏〗^(−𝟏) (𝒙/𝟑) + C

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.