Example 22 - Chapter 7 Class 12 Integrals
Last updated at Dec. 20, 2019 by Teachoo
Transcript
Example 22 Find (i) โซ1โ๐^๐ฅ (tan^(โ1)โก๐ฅ+ 1/(1 + ๐ฅ^2 )) ๐๐ฅ โซ1โใ๐^๐ฅ (tan^(โ1)โก๐ฅ+1/(1 + ๐ฅ^2 ))๐๐ฅใ It is of the form โซ1โใ๐^๐ฅ [๐(๐ฅ)+๐^โฒ (๐ฅ)] ใ ๐๐ฅ=๐^๐ฅ ๐(๐ฅ)+๐ถ Where ๐(๐ฅ)=tan^(โ1)โก๐ฅ ๐^โฒ (๐ฅ)= 1/(1 + ๐ฅ^2 ) So, our equation becomes โซ1โใ๐^๐ฅ (tan^(โ1)โก๐ฅ+1/(1 + ๐ฅ^2 ))๐๐ฅใ=๐^๐ ใ๐ญ๐๐งใ^(โ๐)โกใ๐+๐ชใ Example 22 Find (ii) โซ1โ((๐ฅ^2 + 1) ๐^๐ฅ)/(๐ฅ + 1)^2 ๐๐ฅ โซ1โใ(๐ฅ^2 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ ๐๐ฅใ Adding and subtracting 1 in numerator =โซ1โใ(๐ฅ^2+ 1 + 1 โ 1)/(๐ฅ + 1)^2 .๐^๐ฅ .๐๐ฅใ =โซ1โใ(๐ฅ^2 โ 1 + 1 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ .๐๐ฅใ =โซ1โใ[(๐ฅ^2 โ 1)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ] ๐^๐ฅ ๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ^2 โ (1)^2)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ]๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ โ 1)(๐ฅ + 1)/(๐ฅ + 1)^2 +2/(๐ฅ + 1)^2 ]๐๐ฅใ =โซ1โใ๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)+2/(๐ฅ + 1)^2 ]๐๐ฅใ It is of form โซ1โใ๐^๐ฅ [๐(๐ฅ)+๐^โฒ (๐ฅ)] ใ ๐๐ฅ=๐^๐ฅ ๐(๐ฅ)+๐ถ Where ๐(๐ฅ)=(๐ฅ โ 1)/(๐ฅ + 1) ๐^โฒ (๐ฅ)=๐/๐๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)] ๐^โฒ (๐ฅ)=(1.(๐ฅ + 1) โ1 (๐ฅ โ 1))/(๐ฅ + 1)^2 =(๐ฅ + 1 โ ๐ฅ + 1)/(๐ฅ + 1)^2 =2/(๐ฅ + 1)^2 Thus, our equation becomes โซ1โใ(๐ฅ^2 + 1)/(๐ฅ + 1)^2 .๐^๐ฅ=โซ1โใ๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)+2/(๐ฅ + 1)^2 ]๐๐ฅใใ =๐^๐ฅ [(๐ฅ โ 1)/(๐ฅ + 1)]+๐ถ =(๐ โ ๐)/(๐ + ๐).๐^๐+๐ช
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.