Integration Full Chapter Explained - Integration Class 12 - Everything you need




Last updated at Dec. 20, 2019 by Teachoo
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Example 16 Find โซ1โ(๐ฅ^2+ ๐ฅ +1 ๐๐ฅ )/((๐ฅ + 2) (๐ฅ^2+1) ) We can write equation as (๐ฅ^2+ ๐ฅ + 1)/((๐ฅ + 1) (๐ฅ + 2) )=๐ด/(๐ฅ + 2) + (๐ต๐ฅ + ๐ถ)/(๐ฅ^2+ 1) Cancelling denominator ใ ๐ฅใ^2+ ๐ฅ+1=๐ด(๐ฅ^2+1)+(๐ต๐ฅ+๐ถ) (๐ฅ+2) Putting x = โ๐ (โ2)^2+(โ2)+1=๐ด((โ2)^2+1)+0 4โ2+1= 5A 3/5 = A Putting x = ๐ ๐ฅ^2+ ๐ฅ+1=๐ด(๐ฅ^2+1)+(๐ต๐ฅ+๐ถ) (๐ฅ+2) 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 3/5 + 2C 1 โ 3/5 = 2C 2/5 = 2C C = 1/5 Putting x = 1 ๐ฅ^2+ ๐ฅ+1=๐ด(๐ฅ^2+1)+(๐ต๐ฅ+๐ถ) (๐ฅ+2) 1+1+1= 2A + (B + C)(3) 3 = 2A + 3 (B + C) 3 = 2(3/5) + 3 (B+1/5) 3 โ 6/5 = 3 (B+1/5) 9/5 = 3 (B+1/5) 3/5 โ 1/5 = B B = 2/5 Thus, (๐ฅ^2+ ๐ฅ + 1)/((๐ฅ + 1) (๐ฅ + 2) )=๐ด/(๐ฅ + 2) + (๐ต๐ฅ + ๐ถ)/(๐ฅ^2+ 1) (๐ฅ^2+ ๐ฅ + 1)/((๐ฅ + 1)(๐ฅ^2+ 1)) = 3/(5 (๐ฅ + 2)) + (1 (2๐ฅ + 1))/(5 (๐ฅ^2 + 1)) Hence, our equation becomes โซ1โ(๐ฅ^2+ ๐ฅ + 1)/((๐ฅ + 2) (๐ฅ^2 + 1)) ๐๐ฅ= โซ1โ3/(5(๐ฅ^2 + 1)) ๐๐ฅ+โซ1โ1/5 ((2๐ฅ + 1))/(๐ฅ^2 + 1) ๐๐ฅ = โซ1โ3/(5(๐ฅ^2 + 1)) ๐๐ฅ+ 1/5 โซ1โใ2๐ฅ/(๐ฅ^2 + 1) ๐๐ฅ+ใ 1/5 โซ1โ1/(๐ฅ^2 + 1) ๐๐ฅ ๐๐ 1/5 โซ1โ2๐ฅ/(๐ฅ^2+ 1) ๐๐ฅ Let ๐ก=๐ฅ^2+ 1 ๐๐ก/๐๐ฅ=2๐ฅ ๐๐ก=2๐ฅ ๐๐ฅ Substituting, =1/5 โซ1โ๐๐ก/๐ก = 1/5 log |๐ก| + C_2 = 1/5 log |๐ฅ^2+1| + C_2 ๐๐ 1/5 โซ1โ1/(๐ฅ^2+ 1) ๐๐ฅ = 1/5 ใ๐ก๐๐ใ^(โ1) (๐ฅ)+C_3 Hence โซ1โ(๐ฅ^2+ ๐ฅ + 1)/((๐ฅ + 2) (๐ฅ^2+ 1)) ๐๐ฅ =๐/๐ ๐๐๐|๐+๐|+๐/๐ ๐๐๐|๐^๐+๐|+๐/๐ ใ๐๐๐ใ^(โ๐) (๐)+ C where C = C_1+ C_2+C_3
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