Ex 7.4

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.4, 1 (3๐ฅ^2)/(๐ฅ^6 + 1) We need to find โซ1โ(๐๐^๐)/(๐^๐ + ๐) ๐๐ Let ๐^๐=๐ Diff both sides w.r.t. x 3๐ฅ^2=๐๐ก/๐๐ฅ ๐๐=๐๐/(๐๐^๐ ) Thus, our equation becomes โซ1โ(๐๐^๐)/(๐^๐ + ๐) ๐๐ =โซ1โ(3๐ฅ^2)/((๐ฅ^3 )^2 + 1) ๐๐ฅ Putting the value of ๐ฅ^3=๐ก and ๐๐ฅ=๐๐ก/(3๐ฅ^2 ) =โซ1โ(3๐ฅ^2)/(๐ก^2 + 1) .๐๐ก/(3๐ฅ^2 ) =โซ1โ๐๐ก/(๐ก^2 + 1) =โซ1โ๐๐/(๐^๐ + (๐)^๐ ) =1/1 tan^(โ1)โกใ ๐ก/1 ใ+๐ถ It is of form โซ1โ๐๐ก/(๐ฅ^2 + ๐^2 ) =1/๐ ใใ๐ก๐๐ใ^(โ1) ใโกใ๐ฅ/๐ใ +๐ถ โด Replacing ๐ฅ = ๐ก and ๐ by 1 , we get =tan^(โ1)โกใ (๐ก)ใ+๐ถ =ใใ๐๐๐ใ^(โ๐) ใโก(๐^๐ )+๐ช ("Using" ๐ก=๐ฅ^3 )

Made by

#### 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.