Check sibling questions

Ex 7.5, 19 - Integrate 2x / (x2 + 1) (x2 + 3) - Integration by partial fraction - Type 1

Ex 7.5, 19 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 19 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 19 - Chapter 7 Class 12 Integrals - Part 4


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 + = + + +

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