Check sibling questions

CBSE Class 12 Sample Paper for 2024 Boards

CBSE Class 12 Sample Paper for 2023 Boards →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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 Aug. 10, 2023 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

Slide5.JPG

Slide6.JPG
Slide7.JPG Slide8.JPG

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Book a free demo

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

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.