Check sibling questions

Ex 7.5, 20 - Integrate 1/ x (x4 - 1) - Chapter 7 CBSE - Ex 7.5

Ex 7.5, 20 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 20 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 20 - Chapter 7 Class 12 Integrals - Part 4


Transcript

Ex 7.5, 20 Integrate the function 1/(๐‘ฅ(๐‘ฅ4โˆ’1) ) 1/(๐‘ฅ(๐‘ฅ4 โˆ’ 1) ) Multiplying integrand by ๐‘ฅ^3/๐‘ฅ^3 = 1/(๐‘ฅ(๐‘ฅ^4 โˆ’ 1) ) ร— ๐‘ฅ^3/๐‘ฅ^3 = ๐‘ฅ^3/(๐‘ฅ^4 (๐‘ฅ^4 โˆ’ 1) ) Let t = ๐‘ฅ^4 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 4๐‘ฅ^3 ๐‘‘๐‘ก/(4๐‘ฅ^3 ) = ๐‘‘๐‘ฅ Substituting value of ๐‘ก = ๐‘ฅ^4 & ๐‘‘๐‘ฅ = ๐‘‘๐‘ก/(4๐‘ฅ^3 ) " " โˆซ1โ–’๐‘ฅ^3/(๐‘ฅ^4 (๐‘ฅ^4โˆ’ 1) ) ๐‘‘๐‘ฅ = โˆซ1โ–’๐‘ฅ^3/(๐‘ก(๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก/(4๐‘ฅ^3 ) " " = 1/4 โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก(๐‘ก โˆ’ 1) ) We can write integrand as 1/(๐‘ก(๐‘ก โˆ’ 1) ) = ๐ด/๐‘ก + ๐ต/(๐‘ก โˆ’ 1) 1/(๐‘ก(๐‘ก โˆ’ 1) ) = (๐ด(๐‘ก โˆ’ 1) + ๐ต ๐‘ก)/๐‘ก(๐‘ก โˆ’ 1) Cancelling denominator 1 = ๐ด(๐‘กโˆ’1)+๐ต๐‘ก โ€ฆ(1) Putting t = 0 in (1) 1 = ๐ด(0โˆ’1)+๐ตร—0 1 = ๐ดร—(โˆ’1) 1 = โˆ’๐ด ๐ด = โˆ’1 Putting t = 1 in (1) 1 = A(tโˆ’1)+Bt 1 = ๐ด(1โˆ’1)+๐ตร—1 1 = ๐ดร—0+๐ต 1 = ๐ต ๐ต = 1 Therefore 1/4 โˆซ1โ–’1/(๐‘ก(๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก = โˆซ1โ–’(โˆ’1)/(๐‘ก ) ๐‘‘๐‘ก + โˆซ1โ–’1/(๐‘ก โˆ’ 1 ) = โˆ’ใ€–log ใ€—โก|๐‘ก|+ใ€–log ใ€—โก|๐‘กโˆ’1|+๐ถ = ใ€–log ใ€—โก|(๐‘ก โˆ’ 1)/๐‘ก|+๐ถ Putting back t =ใ€– ๐‘ฅใ€—^4 = ๐Ÿ/๐Ÿ’ ใ€–๐ฅ๐จ๐  ใ€—โก|(๐’™^๐Ÿ’ โˆ’ ๐Ÿ)/๐’™^๐Ÿ’ |+๐‘ช ("As " ๐‘™๐‘œ๐‘” ๐ดโˆ’๐‘™๐‘œ๐‘” ๐ต" = " ๐‘™๐‘œ๐‘” ๐ด/๐ต)

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.