Ex 7.5

Chapter 7 Class 12 Integrals
Serial order wise

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### Transcript

Ex 7.5, 13 2 1 1 + 2 We can write the integrand as 2 1 1 + 2 = 2 1 1 + 2 = 2 1 1 + 2 Let 2 1 1 + 2 = 1 + + 1 + 2 2 1 1 + 2 = 1 + 2 + + 1 1 1 + 2 By cancelling denominator 2 = 1+ 2 + + 1 Putting x = 1 , in (1) 2 = A 1+ 1 2 + Bx + C 1 1 2 = A 1+1 + Bx+C 0 2 = 2A+0 2 2 = A 1 = A A = 1 Putting x = 0 , in (1) 2 = A 1+ 0 2 + B 0 +C 0 1 2 = A 1 +C 1 2 = A C Putting A = 1 2 = 1 C C = 1+2 C = 1 Now, Putting A = 1 , in (1) 2 = A 1+1 + B 1 +C 1 1 2 = 2 + + 2 2 = 2 +2 2 2 = 2+2 2 2+4=2 2 =2 =1 Thus, = 1, =1, = 1 So, we can write 2 1 1 + 2 = 1 + + 1 + 2 = 1 1 + + 1 2 + 1 Therefore integrating 2 1 1 + 2 = 1 1 + + 1 2 + 1 = 1 1 + 2 + 1 + 1 2 + 1 Solving 1 I2 = 2 + 1 Let = 2 +1 = 2 2 = Hence 2 + 1 = . 2 = 2 = 1 2 log + 1 Putting back t = 2 +1 = 1 2 log 2 +1 + 2 Therefore 2 1 1 + 2 = 1 1 + 2 + 1 + 1 2 + 1 = log 1 + 1 2 log 2 +1 + tan 1 + = + + + +

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