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 โˆซ1โ–’ใ€–(๐‘ฅ^2 + 1)/((๐‘ฅ^2 + 2) (๐‘ฅ^2 + 3)) ๐‘‘๐‘ฅใ€— Putting ๐’™^๐Ÿ=๐’š (๐‘ฅ^2 + 1 )/((๐‘ฅ^2 + 2) (๐‘ฅ^2 + 3) )=(๐‘ฆ + 1)/((๐‘ฆ + 2) (๐‘ฆ + 3) ) We can write this in form (๐‘ฆ + 1)/((๐‘ฆ + 2) (๐‘ฆ + 3) )=๐ด/((๐‘ฆ + 2) ) + ๐ต/((๐‘ฆ + 3) ) (๐‘ฆ + 1)/((๐‘ฆ + 2) (๐‘ฆ + 3) )=(๐ด(๐‘ฆ +3) + ๐ต (๐‘ฆ + 2))/((๐‘ฆ + 2) (๐‘ฆ + 3) ) By cancelling denominator ๐‘ฆ+1=๐ด(๐‘ฆ +3) + ๐ต (๐‘ฆ + 2) Putting y = โˆ’3 โˆ’3+1=๐ด(โˆ’3+3)+๐ต(โˆ’3+2) โˆ’2=๐ด ร— 0+๐ต ร— โˆ’1 โˆ’2=โˆ’๐ต ๐‘ฉ=๐Ÿ Putting y = โˆ’2 โˆ’2+1=๐ด(โˆ’2+3)+๐ต(โˆ’2+2) โˆ’1=๐ด ร— 1+๐ต ร— 0 โˆ’1=๐ด ๐‘จ=โˆ’๐Ÿ Hence we can write (๐‘ฆ + 1)/((๐‘ฆ + 2) (๐‘ฆ + 3) )=(โˆ’1)/((๐‘ฆ + 2) ) + 2/((๐‘ฆ + 3) ) Substituting back ๐‘ฆ=๐‘ฅ^2 (๐‘ฅ^2 + 1 )/((๐‘ฅ^2 + 2) (๐‘ฅ^2 + 3) ) =(โˆ’1)/((๐‘ฅ^2 + 2) )+2/((๐‘ฅ^2 + 3) ) Therefore, โˆซ1โ–’(๐‘ฅ^2 + 1 )/((๐‘ฅ^2 + 2) (๐‘ฅ^2 + 3) ) ๐‘‘๐‘ฅ=โˆซ1โ–’(โˆ’1)/((๐‘ฅ^2 + 2) ) ๐‘‘๐‘ฅ+โˆซ1โ–’2/((๐‘ฅ^2 + 3) ) ๐‘‘๐‘ฅ =โˆ’โˆซ1โ–’1/((๐‘ฅ^2 +(โˆš2)^2 ) ) ๐‘‘๐‘ฅ+2โˆซ1โ–’1/((๐‘ฅ^2 +(โˆš3)^2 ) ) ๐‘‘๐‘ฅ By using formula โˆซ1โ–’1/(๐‘ฅ^2 + ๐‘Ž^2 ) ๐‘‘๐‘ฅ=1/๐‘Ž ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1)โก(๐‘ฅ/๐‘Ž)+๐ถ =(โˆ’๐Ÿ)/โˆš๐Ÿ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐’™/โˆš๐Ÿใ€—+๐Ÿ/โˆš๐Ÿ‘ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โกใ€–๐’™/โˆš๐Ÿ‘ใ€— +๐‘ช

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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.