Ex 7.5, 19 - Chapter 7 Class 12 Integrals
Last updated at July 11, 2018 by Teachoo
Transcript
Ex 7.5, 19 2 ( 2 + 1)( 2 + 3) Let 2 = Differentiate both sides . . . . 2 = = 2 Substituting value of & 2 ( 2 + 1)( 2 + 3) = 2 + 1 + 3 = 2 + 1 + 3 2 = + 1 + 3 Now we can write 1 + 1 + 3 = + 1 + + 3 1 + 1 + 3 = + 3 + + 1 + 1 + 3 Cancelling denominator 1 = +3 + +1 Putting t = 1 in (1) 1 = 1+3 + 1+1 1 = 2+ 0 1 = 2 = 1 2 Similarly Putting t= 3 in (1) 1 = +3 + B ( +1) 1 = 3+3 + 3+1 1 = 0+ 2 1 = 2 = 1 2 Therefore 1 + 1 + 3 = 1 2 + 1 + 1 2 + 3 = 1 2 log +1 1 2 log +3 + = 1 2 log [ +1 log |t+3|]+C = 1 2 log + 1 + 3 + Substituting back the value of t = 2 = 1 2 log 2 + 1 2 + 3 + = + + +
Ex 7.5
Ex 7.5, 1
Ex 7.5, 2
Ex 7.5, 3
Ex 7.5, 4
Ex 7.5, 5
Ex 7.5, 6
Ex 7.5, 7
Ex 7.5, 8
Ex 7.5, 9 Important
Ex 7.5, 10
Ex 7.5, 11 Important
Ex 7.5, 12
Ex 7.5, 13
Ex 7.5, 14
Ex 7.5, 15
Ex 7.5, 16
Ex 7.5, 17 Important
Ex 7.5, 18 Important
Ex 7.5, 19 You are here
Ex 7.5, 20
Ex 7.5, 21 Important
Ex 7.5, 22
Ex 7.5, 23
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.