Ex 7.2

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.2, 19 Integrate the function (๐2๐ฅ โ 1)/(๐2๐ฅ+ 1) Simplify the given function (๐^2๐ฅ โ 1)/(๐^2๐ฅ + 1) Dividing numerator and denominator by ex, we obtain = (๐^2๐ฅ/๐^๐ฅ " " โ" " ๐/๐^๐ )/(๐^2๐ฅ/๐^๐ฅ " " + " " ๐/๐^๐ ) = (๐^๐ โ ๐^(โ๐))/(๐^๐ + ๐^(โ๐) ) Let ๐^๐ฅ + ๐^(โ๐ฅ)= ๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐^๐ฅ+(โ1) ๐^(โ๐ฅ)= ๐๐ก/๐๐ฅ ๐^๐ฅโ๐^(โ๐ฅ)= ๐๐ก/๐๐ฅ ๐๐ฅ=๐๐ก/(๐^๐ฅ โ ๐^(โ๐ฅ) ) Now, Integrating the function โซ1โใ" " (๐^2๐ฅ โ 1)/(๐^2๐ฅ + 1) " " ใ. ๐๐ฅ = โซ1โใ" " (๐^๐ฅ โ ๐^(โ๐ฅ))/(๐^๐ฅ + ๐^(โ๐ฅ) ) " " ใ. ๐๐ฅ (Using (๐^2๐ฅ โ 1)/(๐^2๐ฅ + 1)=(๐^๐ฅ โ ๐^(โ๐ฅ))/(๐^๐ฅ + ๐^(โ๐ฅ) ) ) Putting ๐^๐ฅ + ๐^(โ๐ฅ)=๐ก & ๐๐ฅ=๐๐ก/(๐^๐ฅ โ ๐^(โ๐ฅ) ) =โซ1โใ" " (๐^๐ฅ โ ๐^(โ๐ฅ))/๐ก " " ใ. ๐๐ก/(๐^๐ฅ โ ๐^(โ๐ฅ) ) " " =โซ1โใ" " 1/๐ก " " ใ. ๐๐ก =logโก|๐ก|+๐ถ =logโกใ |๐^๐ฅ+๐^(โ๐ฅ) |ใ+๐ถ =๐๐๐โก(๐^๐ + ๐^(โ๐) )+๐ช (Using ๐ก=๐^๐ฅ + ๐^(โ๐ฅ)) (As ๐^๐ฅ+๐^(โ๐ฅ)>0 )

#### 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.