Ex 7.7

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.7, 13 (Supplementary NCERT) ๐ฅ+1 โ(2๐ฅ^2+3) ๐ฅ+1 โ( 2๐ฅ^2+3) Integrating the function w.r.t.x โซ1โใ(๐ฅ+1) โ(ใ2๐ฅใ^2+3)ใ ๐๐ฅ = โซ1โใ[๐ฅโ(ใ2๐ฅใ^2+3)+1โ(ใ2๐ฅใ^2+3)] ๐๐ฅใ = โซ1โใ๐ฅ โ(ใ2๐ฅใ^2+3) ๐๐ฅ+โซ1โใโ(ใ2๐ฅใ^2+3 ) ๐๐ฅใใ Solving ๐_๐ ๐_๐=โซ1โใ๐ฅ โ(ใ2๐ฅใ^2+3)ใ ๐๐ฅ Let ใ2๐ฅใ^2+3=๐ก Diff. both sides w.r.t.x 4x + 0 = ๐๐ก/๐๐ฅ dx = ๐๐ก/4๐ฅ Now, equation becomes I_1 = โซ1โใ๐ฅ โ(ใ2๐ฅใ^2+3)ใ ๐๐ฅ Putting the value of (ใ2๐ฅใ^2+3) and dx I_1 = โซ1โใ๐ฅ โ๐กใ . ๐๐ฅ I_1 = โซ1โใ๐ฅ โ๐กใ ๐๐ก/4๐ฅ I_1 = 1/4 โซ1โใ โ๐ก ใ. ๐๐ก I_1 = 1/4 โซ1โใ ๐ก^(1/2) ใ . ๐๐ก I_1 = 1/4 [ใ๐ก ใ^(1/2 + 1)/((1/2 + 1) )]+C_1 I_1 = 1/4 [ใ๐ก ใ^(3/2 )/(3/2)]+C_1 I_1 = ใ๐ก ใ^(3/2 )/6+C_1 I_1 = (ใ2๐ฅใ^2+3)^(3/2)/6+C_1 Solving ๐ฐ_๐ I_2 = โซ1โโ(ใ2๐ฅใ^2+3) ๐๐ฅ I_2 = โซ1โโ(2 (ใ ๐ฅใ^2+3/2) ) ๐๐ฅ I_2 = โซ1โใโ2 โ( ใ ๐ฅใ^2+3/2)ใ ๐๐ฅ I_2 =โ2 โซ1โโ(ใ ๐ฅใ^2+3/2) ๐๐ฅ I_2 =โ2 โซ1โโ(ใ ๐ฅใ^2+(โ(3/2))^2 ) ๐๐ฅ It is of form โซ1โใโ(๐ฅ^2+๐^2 ) ใ ๐๐ฅ=1/2 ๐ฅ โ(๐ฅ^2+๐^2 )+๐^2/2 logโก|๐ฅ+โ(๐ฅ^2+๐^2 )|+C_2 Replacing a by โ(3/2), we get I_2 =โ2 [๐ฅ/2 โ(ใ ๐ฅใ^2+(โ(3/2))^2 )]+โ2 ใ((โ(3/2 )))/2ใ^2 logโก|๐ฅ+โ(ใ ๐ฅใ^2+(โ(3/2))^2 )|+C_2 I_2 =โ2 [๐ฅ/2 โ(ใ ๐ฅใ^2+3/2)]+โ2 (3/2)/2 logโก|๐ฅ+โ(ใ ๐ฅใ^2+3/2)|+C_2 I_2 =(โ2 ๐ฅ)/2.โ(ใ 2๐ฅใ^2+3)/โ2+(3โ2)/4 logโก|๐ฅ+โ(ใ ๐ฅใ^2+3/2)|+ใโ2 Cใ_2 I_2 =๐ฅ/2.โ(ใ 2๐ฅใ^2+3)+(3โ2)/4 logโก|๐ฅ+โ(ใ ๐ฅใ^2+3/2)|+ใ Cใ_3 Now, put the value of I_1 and I_2 in eq. (1), we get โซ1โใ(๐ฅ+1) โ(ใ2๐ฅใ^2+3)ใ ๐๐ฅ=โซ1โ๐ฅ โ(ใ2๐ฅใ^2+3) ๐๐ฅ+โซ1โโ(ใ2๐ฅใ^2+3) ๐๐ฅ = (ใ2๐ฅใ^2+3)^(3/2)/6+ใ Cใ_1+๐ฅ/2 โ(ใ2๐ฅใ^2+3)+(3โ2)/4 logโก|๐ฅ+โ(ใ ๐ฅใ^2+3/2)|+ใ Cใ_3 = (ใ๐๐ใ^๐+๐)^(๐/๐)/๐+๐/๐ โ(ใ๐๐ใ^๐+๐)+(๐โ๐)/๐ ๐๐๐โก|๐+โ(ใ ๐ใ^๐+๐/๐)|+๐