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




Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.9, 16 โซ_1^2โ(5๐ฅ^2)/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅ Let F(๐ฅ)=โซ1โใ(5๐ฅ^2)/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅใ =5โซ1โใ๐ฅ^2/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅใ =5โซ1โใ(๐ฅ^2 + 4๐ฅ + 3 โ 4๐ฅ โ 3)/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅใ =5โซ1โใ(๐ฅ^2 + 4๐ฅ + 3)/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅใโ5โซ1โใ( (4๐ฅ + 3))/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅใ =โซ1โใ(5โ(20๐ฅ + 15 )/(๐ฅ^2 + 4๐ฅ + 3)) ๐๐ฅใ =โซ1โใ(5โ(20๐ฅ + 15 )/(๐ฅ^2 + 3๐ฅ + ๐ฅ + 3)) ๐๐ฅใ =โซ1โใ(5โ(20๐ฅ + 15 )/(๐ฅ(๐ฅ + 3) + 1(๐ฅ + 3))) ๐๐ฅใ =โซ1โใ(5โ(20๐ฅ + 15 )/((๐ฅ + 3) (๐ฅ + 1))) ๐๐ฅใ Now, Let (20๐ฅ + 15)/(๐ฅ + 3)(๐ฅ + 1) =A/(๐ฅ + 3)+B/(๐ฅ + 1) (20๐ฅ + 15)/(๐ฅ + 3)(๐ฅ + 1) =(A(๐ฅ + 1) + B(๐ฅ + 3))/(๐ฅ + 3)(๐ฅ + 1) Canceling denominator 20๐ฅ+15=A(๐ฅ + 1) + B(๐ฅ + 3) Putting ๐ฅ=โ1 20(โ1)+15=A(โ1 + 1) + B(โ1 + 3) โ20+15=Aร0+B (2) Putting ๐ฅ=โ1 20(โ1)+15=A(โ1 + 1) + B(โ1 + 3) โ20+15=Aร0+B (2) โ5=2B B=(โ5)/( 2) Similarly Putting ๐ฅ=โ3 20(โ3)+15=A(โ3+1)+B(โ3+3) โ60+15=A(โ2) Bร0 โ45=โ2A A=45/2 Hence โซ1โโ((5๐ฅ^2)/(๐ฅ^2 + 4๐ฅ + 3) " " =โซ1โใ(5โ(20๐ฅ + 15 )/(๐ฅ^2 + 4๐ฅ + 3)) ๐๐ฅใ) =โซ1โใ5โA/(๐ฅ+3)โใ B/(๐ฅ+1) ๐๐ฅ =โซ1โใ5 ๐๐ฅใโโซ1โใ(45/2)/(๐ฅ +_3) ๐๐ฅโโซ1โใ(((โ5)/( 2)))/(๐ฅ+1) ๐๐ฅใใ =5๐ฅโ45/2 ๐๐๐|๐ฅ+3|+5/2 ๐๐๐|๐ฅ+1| Hence F(๐ฅ)=5๐ฅโ5/2 [9 ๐๐๐|๐ฅ+3|โ๐๐๐|๐ฅ+1|] Now, โซ_1^2โใ(5๐ฅ^2)/(๐ฅ^2 + 4๐ฅ + 3) ๐๐ฅ=๐น(2)โ๐น(1) ใ =[5ร2โ5/2 (9 ๐๐๐|2+3|โ๐๐๐|2+1|)] โ [5 ร1โ5/2 (9 ๐๐๐|1+3|โ๐๐๐|1+1|)] =10โ5/2 [9 ๐๐๐ 5โ๐๐๐ 3]โ5+5/2 [9 ๐๐๐ 4โ๐๐๐ 2] =10โ5โ5/2 [9๐๐๐ 5โ๐๐๐ 3โ9๐๐๐ 4+๐๐๐ 2)] =10โ5โ5/2 [9๐๐๐ 5โ9๐๐๐ 4โ(๐๐๐ 3โ๐๐๐ 2)] =๐โ๐/๐ (๐ ๐ฅ๐จ๐ ๐/๐โ๐ฅ๐จ๐ ๐/๐) (log a โ log b = log ๐/๐)
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