Ex 7.7

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.7, 14 (Supplementary NCERT) (๐ฅ+3) โ(3โ4๐ฅใโ๐ฅใ^2 ) (๐ฅ+3) โ(3โ4๐ฅใโ๐ฅใ^2 ) We can write it as:- x + 3 = A [๐/๐๐ฅ (3โ4๐ฅโ๐ฅ^2 )]+ B x + 3 = A [0โ4โ2๐ฅ]+ B x + 3 = A [โ4โ2๐ฅ]+ B x + 3 = โ4"A"โ2"A" ๐ฅ+ B x + 3 = โ2"A" ๐ฅ+(โ4"A"+๐ต) Comparing x and constant term Thus, we can write x + 3 = A [โ4โ2๐ฅ] + B x + 3 = (โ1)/2 [โ4โ2๐ฅ] + 1 x = (โ2A)x ๐ฅ/๐ฅ = โ2A 1 = โ2A A = (โ1)/2 3 = โ4A + B 3 = โ4((โ1)/2) + B 3 = 2 + B B = 3 โ 2 B = 1 Integrating the function w.r.t.x โซ1โใ(๐ฅ+3) โ(3โ4๐ฅโ๐ฅ^2 ) ใ ๐๐ฅ = โซ1โใ[โ1/2 [โ4 โ2๐ฅ]+1] ใ โ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ = โซ1โใ[โ1/2 [โ4 โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 )+1โ(3โ4๐ฅโ๐ฅ^2 )] ใ ๐๐ฅ = โซ1โใโ1/2 [โ4 โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ+ใ โซ1โโ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ = โ1/2 โซ1โใ[โ4 โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ+ใ โซ1โโ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ Solving ๐ฐ_๐ I_1 = (โ1)/2 โซ1โใ[โ4โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 )ใ ๐๐ฅ Let 3 โ 4๐ฅ โ ๐ฅ^2 = t Diff. both sides w.r.t.x 0 โ 4 โ2x = ๐๐ก/๐๐ฅ โ 4 โ 2x = ๐๐ก/๐๐ฅ dx = ๐๐ก/(โ4 โ 2๐ฅ) Thus, our equation becomes I_1 = (โ1)/2 โซ1โใ[โ4โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 )ใ ๐๐ฅ Putting the value if (3โ4๐ฅโ๐ฅ^2) and dx, we get I_1 = (โ1)/2 โซ1โใ[โ4โ2๐ฅ] โ๐กใ. ๐๐ฅ I_1 = (โ1)/2 โซ1โใ[โ4โ2๐ฅ] โ๐กใ. ๐๐ก/[โ4โ2๐ฅ] ("Using t = " 3โ4๐ฅ โ๐ฅ^2 ) Solving ๐ฐ_๐ I_2 = โซ1โโ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ I_2 = โซ1โโ(3โ(4๐ฅ+๐ฅ^2)) ๐๐ฅ I_2 = โซ1โโ(3โ(๐ฅ^2+4๐ฅ)) ๐๐ฅ I_2 = โซ1โใโ(3โ[๐ฅ^2+2(2) (๐ฅ))] ใ ๐๐ฅ I_2 = โซ1โใโ(3โ[๐ฅ^2+2(2)+(2)^2 โใ(2)ใ^2 )] ใ ๐๐ฅ I_2 = โซ1โใโ(3โ[(ใ๐ฅ+2)ใ^2โ ใ(2)ใ^2 )] ใ ๐๐ฅ I_2 = โซ1โใโ(3โ(๐ฅ+2)^2+ ใ(2)ใ^2 ) ใ ๐๐ฅ I_2 = โซ1โใโ(3โ(๐ฅ+2)^2+4) ใ ๐๐ฅ I_2 = โซ1โใโ(7โ(๐ฅ+2)^2 ) ใ ๐๐ฅ I_2 = โซ1โใโ(ใ(โ7)ใ^2โ(๐ฅ+2)^2 ) ใ ๐๐ฅ I_2 = (๐ฅ + 2)/2 โ((โ7)^2โ(๐ฅ+2)^2 )+ใ(7)ใ^2/2 ใ๐ ๐๐ใ^(โ1) ((๐ฅ + 2)/โ7)+ C_2 I_2 = (๐ฅ + 2)/2 โ(7โใ(๐ฅใ^2 + 4๐ฅ +4)) +7/2 ใ๐ ๐๐ใ^(โ1) ((๐ฅ + 2)/โ7)+ C_2 I_2 = (๐ฅ + 2)/2 โ(7โ๐ฅ^2 + 4๐ฅ +4) +7/2 ใ๐ ๐๐ใ^(โ1) ((๐ฅ + 2)/โ7)+ C_2 I_2 = (๐ฅ + 2)/2 โ(3โ 4๐ฅโ๐ฅ^2 ) +7/2 ใ๐ ๐๐ใ^(โ1) ((๐ฅ + 2)/โ7)+ C_2 It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=1/2 ๐ฅโ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1) (๐ฅ/๐)+ ๐ถ_2 Replacing x by (x + 2) a by โ7 , we get Putting the value of I_1 and I_2 in (1) โซ1โ(๐ฅ+3) โ(3โ 4๐ฅโ๐ฅ^2 ) d๐ฅ = (โ1)/2 โซ1โใ[โ4โ2๐ฅ] โ(3โ4๐ฅโ๐ฅ^2 )ใ ๐๐ฅ+โซ1โโ(3โ4๐ฅโ๐ฅ^2 ) ๐๐ฅ = (โ1)/3 ใ(3โ4๐ฅโ๐ฅ^2)ใ^(3/2) + C_1+ ((๐ฅ +2) โ(3 โ 4๐ฅ โ ๐ฅ^2 ))/2+7/2 ใ๐ ๐๐ใ^(โ1) ((๐ฅ + 2)/โ7)+ C_3 = (โ๐)/๐ ใ(๐โ๐๐โ๐^๐)ใ^(๐/๐) +๐/๐ ใ๐๐๐ใ^(โ๐) ((๐ + ๐)/โ๐)+ ((๐ +๐) โ(๐ โ ๐๐ โ ๐^๐ ))/๐+ ๐