Misc 23 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Integration by substitution - x^n
Last updated at April 16, 2024 by Teachoo
Misc 23 Integrate the function (โ(๐ฅ^2 + 1) [logโกใ(๐ฅ^2+ 1) โ 2 logโก๐ฅ ใ ] )/๐ฅ^4 โซ1โ(โ(๐ฅ^2 + 1) [logโกใ(๐ฅ^2+ 1) โ 2 logโก๐ฅ ใ ] )/๐ฅ^4 ๐๐ฅ Taking ๐ฅ^2common from โ(๐ฅ^2+1) = โซ1โ(ใใ(๐ฅใ^2) ใ^(1/2) (1 + 1/๐ฅ^2 )^(1/2) (logโกใ(๐ฅ^2+1)ใ โ logโกใ๐ฅ^2 ใ ))/๐ฅ^4 ๐๐ฅ = โซ1โ(๐ฅ (1+ 1/๐ฅ^2 )^(1/2) (logโกใ ((๐ฅ^(2 )+ 1))/๐ฅ^2 ใ ))/๐ฅ^4 ๐๐ฅ = โซ1โ( (1+ 1/๐ฅ^2 )^(1/2) (logโก(1 + 1/๐ฅ^2 ) ))/๐ฅ^3 Let t = 1 + 1/๐ฅ^2 ๐๐ก/๐๐ฅ=(โ2)/๐ฅ^3 (โ1)/2 ๐๐ก=๐๐ฅ/๐ฅ^3 Substituting, = โ1/2 โซ1โ๐ก^(1/2) ใ log ๐กใโกใ ๐๐กใ = โซ1โ( (1+ 1/๐ฅ^2 )^(1/2) (logโก(1 + 1/๐ฅ^2 ) ))/๐ฅ^3 Let t = 1 + 1/๐ฅ^2 ๐๐ก/๐๐ฅ=(โ2)/๐ฅ^3 (โ1)/2 ๐๐ก=๐๐ฅ/๐ฅ^3 Substituting value of t and dt = (โ1)/2 โซ1โ๐ก^(1/2) ใ log ๐กใโกใ ๐๐กใ Hence, (โ1)/2 โซ1โใ๐ก^(1/2) logโกใ๐ก ๐๐ก=(โ1)/2 (logโกใ๐ก โซ1โใ๐ก^(1/2) ๐๐กใโโซ1โ((๐(logโกใ๐ก)ใ)/๐๐ก โซ1โ๐ก^(1/2) ๐๐ก) ๐๐กใ )ใ ใ = (โ1)/2 (logโกใ๐ก (๐ก^(3/2)/(3/2))โโซ1โใ1/๐กร(๐ก^(3/2)/(3/2)) ใใ ๐๐ก) = (โ1)/2 (2/3 ๐ก^(3/2) logโกใ๐กโ2/3ใ โซ1โใ๐ก^(1/2) ๐๐กใ) = (โ1)/2 (2/3 ๐ก^(3/2) logโกใ๐กโ2/3ใ ( ใ2๐กใ^(3/2))/3) = (โ1)/3 ๐ก^(3/2) logโก๐ก + 2/9 ๐ก^(3/2) Putting value of t = 1 + 1/๐ฅ^2 = (โ1)/3 (1+1/๐ฅ^2 )^(3/2) logโกใ(1+1/๐ฅ^2 )+2/9 " " (1+1/๐ฅ^2 )^(3/2)+ใ C = (โ๐)/๐ (๐+๐/๐^(๐ ) )^(๐/๐) (๐ฅ๐จ๐ โก(๐+๐/๐^๐ )โ๐/๐)+ C