Integration by substitution - e^x

Chapter 7 Class 12 Integrals
Concept 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 )