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
CBSE Class 12 Sample Paper for 2024 Boards
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CBSE Class 12 Sample Paper for 2024 Boards
Last updated at April 16, 2024 by Teachoo
(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