Integration by partial fraction - Type 3

Chapter 7 Class 12 Integrals
Concept wise

### Transcript

Ex 7.5, 10 2 3 2 1 2 + 3 We can write the integrand as 2 3 2 1 2 + 3 = 2 3 1 + 1 2 + 3 = + 1 + 1 + 2 + 3 = 1 2 + 3 + + 1 2 + 3 + 1 + 3 1 + 1 2 + 3 By cancelling denominator 2 3 = 1 2 +3 + +1 2 +3 + 1 +1 By substituting x = 1, in (1) 2 3 = 1 1 2+3 + 1+1 2+3 + 1+1 1 1 5 = 2 1 + 0+ 0 5 = 2 = 5 2 Similarly Putting x = 1, in (1) 2 3 = 1 1 2+3 + 1+1 2+3 + 1 1 1+1 1 = A 0 + 2 5 + 0 1 = 10 = 1 10 Similarly Putting x = 3 2 , in (1) 2 3 2 3 = 3 2 1 2 3 2 +3 + 3 2 +1 2 3 2 +3 + 3 2 1 3 2 +1 6 = 0+ 0+ 1 2 5 2 6 = 5 4 = 24 5 Therefore 2 3 + 1 1 2 + 3 = 5 2 + 1 + 1 10 1 + 24 5 2 + 3 Hence we can write it as 2 3 + 1 1 2 + 3 = 5 2 + 1 + 1 10 1 + 24 5 2 + 3 = 5 2 log +1 1 10 log 1 24 5 1 2 log 2 +3 + = + + +

#### 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.