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Ex 7.5, 13 - Integrate 2 / (1 - x) (1 + x2) - Chapter 7 NCERT - Integration by partial fraction - Type 5

Ex 7.5, 13 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 13 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 13 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.5, 13 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.5, 13 - Chapter 7 Class 12 Integrals - Part 6


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 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.