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Find ∫ x 2 + 1 / (x 2 + 2) (x 2 + 3) dx

Find Integration ∫ (x^2 + 1)/ (x^2 + 2) (x^2 + 3) dx - Teachoo Maths

Question 33 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 33 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

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Question 33 Find ∫1β–’γ€–(π‘₯^2 + 1)/((π‘₯^2 + 2) (π‘₯^2 + 3)) 𝑑π‘₯γ€— Putting 𝒙^𝟐=π’š (π‘₯^2 + 1 )/((π‘₯^2 + 2) (π‘₯^2 + 3) )=(𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) ) We can write this in form (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=𝐴/((𝑦 + 2) ) + 𝐡/((𝑦 + 3) ) (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=(𝐴(𝑦 +3) + 𝐡 (𝑦 + 2))/((𝑦 + 2) (𝑦 + 3) ) By cancelling denominator 𝑦+1=𝐴(𝑦 +3) + 𝐡 (𝑦 + 2) Putting y = βˆ’3 βˆ’3+1=𝐴(βˆ’3+3)+𝐡(βˆ’3+2) βˆ’2=𝐴 Γ— 0+𝐡 Γ— βˆ’1 βˆ’2=βˆ’π΅ 𝑩=𝟐 Putting y = βˆ’2 βˆ’2+1=𝐴(βˆ’2+3)+𝐡(βˆ’2+2) βˆ’1=𝐴 Γ— 1+𝐡 Γ— 0 βˆ’1=𝐴 𝑨=βˆ’πŸ Hence we can write (𝑦 + 1)/((𝑦 + 2) (𝑦 + 3) )=(βˆ’1)/((𝑦 + 2) ) + 2/((𝑦 + 3) ) Substituting back 𝑦=π‘₯^2 (π‘₯^2 + 1 )/((π‘₯^2 + 2) (π‘₯^2 + 3) ) =(βˆ’1)/((π‘₯^2 + 2) )+2/((π‘₯^2 + 3) ) Therefore, ∫1β–’(π‘₯^2 + 1 )/((π‘₯^2 + 2) (π‘₯^2 + 3) ) 𝑑π‘₯=∫1β–’(βˆ’1)/((π‘₯^2 + 2) ) 𝑑π‘₯+∫1β–’2/((π‘₯^2 + 3) ) 𝑑π‘₯ =βˆ’βˆ«1β–’1/((π‘₯^2 +(√2)^2 ) ) 𝑑π‘₯+2∫1β–’1/((π‘₯^2 +(√3)^2 ) ) 𝑑π‘₯ By using formula ∫1β–’1/(π‘₯^2 + π‘Ž^2 ) 𝑑π‘₯=1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑(π‘₯/π‘Ž)+𝐢 =(βˆ’πŸ)/√𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖𝒙/βˆšπŸγ€—+𝟐/βˆšπŸ‘ 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖𝒙/βˆšπŸ‘γ€— +π‘ͺ

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.