Check sibling questions

Ex 7.5, 2 - Integrate 1 / x2 - 9 - Chapter 7 Class 12 NCERT - Ex 7.5

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


Transcript

Ex 7.5, 2 1/(๐‘ฅ2โˆ’ 9) Solving integrand 1/(๐‘ฅ2โˆ’ 9)=1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) We can write it as 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) )=๐ด/((๐‘ฅ โˆ’ 3) ) + ๐ต/((๐‘ฅ + 3) ) 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) )=(๐ด(๐‘ฅ + 3) + ๐ต(๐‘ฅ โˆ’ 3))/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) " " Cancelling denominator 1 = ๐ด(๐‘ฅ + 3) + ๐ต(๐‘ฅ โˆ’ 3) Putting ๐‘ฅ = 3 in (1) 1 = ๐ด(๐‘ฅ+3)+๐ต(๐‘ฅโˆ’3) 1 = ๐ด(3+3) + ๐ต(3โˆ’3) 1 = ๐ดร—6+ ๐ตร—0 1 = 6๐ด ๐ด = 1/6 Similarly Putting y=โˆ’3 in (1) 1 = ๐ด(โˆ’3+3) + ๐ต(โˆ’3โˆ’3) 1 = ๐ดร—0+ ๐ตร—(โˆ’6) 1 = โˆ’6๐ต ๐ต = (โˆ’1)/6 Hence we can write it as 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) = 1/6(๐‘ฅ โˆ’ 3) โˆ’ 1/6(๐‘ฅ + 3) Therefore โˆซ1โ–’1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) ๐‘‘๐‘ฅ = 1/6 โˆซ1โ–’1/((๐‘ฅ โˆ’ 3) ) ๐‘‘๐‘ฅ โˆ’ 1/6 โˆซ1โ–’1/((๐‘ฅ + 3) ) ๐‘‘๐‘ฅ = 1/6 ใ€–log ใ€—โก|๐‘ฅโˆ’3|โˆ’ 1/6 ใ€–log ใ€—โก|๐‘ฅ+3|+๐ถ = 1/6 (ใ€–log ใ€—โก|๐‘ฅโˆ’3|โˆ’ใ€–log ใ€—โก|๐‘ฅ+3| )+๐ถ = ๐Ÿ/๐Ÿ” ใ€–๐’๐’๐’ˆ ใ€—โก|(๐’™ โˆ’ ๐Ÿ‘)/(๐’™ + ๐Ÿ‘)|+๐‘ช

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.