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  1. Chapter 7 Class 12 Integrals
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Ex 7.7, 12 (Supplementary NCERT) ๐‘ฅ โˆš(๐‘ฅ+ ๐‘ฅ^2 ) ๐‘ฅ โˆš(๐‘ฅ+ ๐‘ฅ^2 ) Step 1: It can be written in the form x = A [๐‘‘/๐‘‘๐‘ฅ (๐‘ฅ+๐‘ฅ^2 )]+ B x = A [1+2๐‘ฅ]+ B x = (2A) x + A +B x = (2A) x ๐‘ฅ/๐‘ฅ = 2A 1 = 2A A = 1/20 = A + B B = โˆ’ A B = (โˆ’1)/2 Thus, x = A [1 + 2x] + B x = 1/2 [1 + 2x] โˆ’ 1/2 Step 2: Integrating the function w.r.t. x โˆซ1โ–’ใ€–๐‘ฅโˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[1/2 [1+2๐‘ฅ]โˆ’1/2] โˆš(๐‘ฅ+๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–1/2 [1+2๐‘ฅ] โˆš(๐‘ฅ+๐‘ฅ^2 ) ใ€— ๐‘‘๐‘ฅโˆ’โˆซ1โ–’ใ€–1/2 โˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = 1/2 โˆซ1โ–’(1+2๐‘ฅ) โˆš(๐‘ฅ+๐‘ฅ^2 ) ๐‘‘๐‘ฅ โˆ’1/2 โˆซ1โ–’โˆš(๐‘ฅ+๐‘ฅ^2 ) ๐‘‘๐‘ฅ Solving ๐ˆ_๐Ÿ ๐ผ_1= 1/2 โˆซ1โ–’ใ€–(1+2๐‘ฅ) โˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— . ๐‘‘๐‘ฅ Let ๐‘ฅ + ๐‘ฅ^2=๐‘ก Diff. both sides w.r.t.x 1 + 2๐‘ฅ=๐‘‘๐‘ก/๐‘‘๐‘ฅ dx = ๐‘‘๐‘ก/(1 + 2๐‘ฅ) Thus, our equation becomes As I_1= 1/2 โˆซ1โ–’ใ€–(1+2๐‘ฅ) โˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— .๐‘‘๐‘ฅ Putting the value of (๐‘ฅ+๐‘ฅ^2) and d๐‘ฅ I_1= 1/2 โˆซ1โ–’ใ€–(1+2๐‘ฅ) โˆš๐‘กใ€— .๐‘‘๐‘ก/((1 + 2๐‘ฅ) ) I_1= 1/2 โˆซ1โ–’โˆš๐‘ก ๐‘‘๐‘ก I_1= 1/2 โˆซ1โ–’(๐‘ก)^(1/2) ๐‘‘๐‘ก I_1= 1/2 ๐‘ก^(ใ€–1/2ใ€—^(+ 1) )/((1/2 + 1) ) + C_1 I_1= 1/2 ๐‘ก^((1 + 2)/2)/(((1 + 2)/2 ) ) + C_1 I_1= 1/2 ๐‘ก^(3/2)/(3/2) + C_1 I_1= ๐‘ก^(3/2)/3 + C_1 I_1= ใ€–(๐‘ฅ^2 + ๐‘ฅ)ใ€—^(3/2)/3 + C_1 Solving ๐‘ฐ_๐Ÿ I_2= 1/2 โˆซ1โ–’โˆš(๐‘ฅ+๐‘ฅ^2 ) ๐‘‘๐‘ก I_2 = 1/2 โˆซ1โ–’โˆš(๐‘ฅ^2+๐‘ฅ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(๐‘ฅ^2+2(๐‘ฅ)(1/2) ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš(๐‘ฅ^2+2(๐‘ฅ)(1/2)+(1/2)^2โˆ’(1/2)^2 ) ๐‘‘๐‘ฅ I_2 = 1/2 โˆซ1โ–’โˆš((๐‘ฅ+1/2)^2โˆ’(1/2)^2 ) ๐‘‘๐‘ฅ It is of form โˆซ1โ–’ใ€–โˆš(๐‘ฅ^2โˆ’๐‘Ž^2 ) ใ€— ๐‘‘๐‘ฅ=๐‘ฅ/2 โˆš(๐‘ฅ^2 โˆ’ ๐‘Ž^2 )โˆ’๐‘Ž^2/2 logโก|๐‘ฅ+โˆš(๐‘ฅ^2โˆ’๐‘Ž^2 )|+C_2 Replacing x by (x + 1/2 ) and a by 1/2, we get I_2 = 1/2 [(๐‘ฅ + 1/2)/2 โˆš((๐‘ฅ+1/2)^2โˆ’(1/2)^2 )โˆ’(1/2)^2/2 log |๐‘ฅ+1/2+โˆš((๐‘ฅ+1/2)^2โˆ’(1/2)^2 )|+ C_2 ] I_2 = 1/2 [((2๐‘ฅ+1)/2)/2 โˆš((๐‘ฅ+1/2)^2โˆ’(1/2)^2 )โˆ’(1/2)^2/2 log|๐‘ฅ+1/2+โˆš((๐‘ฅ+1/2)^2โˆ’(1/2)^2 )|+ C_2 ] I_2 = 1/2 [(2๐‘ฅ+1)/2.2 โˆš(๐‘ฅ^2โˆ’(1/2)^2+2(๐‘ฅ)(1/2)โˆ’(1/2)^2 )โˆ’(1/4)/2 log โ”œ|๐‘ฅ+1/2+โˆš((๐‘ฅ+1/2)^2+2(๐‘ฅ)(1/2) )โ”ค+ C_2 ] I_2 = 1/2 [(2๐‘ฅ+1)/4 โˆš(๐‘ฅ^2+๐‘ฅ)โˆ’1/(4 . 2 ) log โ”œ|๐‘ฅ+1/2+โˆš((๐‘ฅ)^2+2(๐‘ฅ)(1/2) )โ”ค+ C_2 ] I_2 = 1/2 [(2๐‘ฅ+1)/4 โˆš(๐‘ฅ^2+๐‘ฅ)โˆ’1/8 log |๐‘ฅ+1/2+โˆš((๐‘ฅ)^2+๐‘ฅ)|+ C_2 ] I_2 = ((2๐‘ฅ + 1))/8 โˆš(๐‘ฅ^2+๐‘ฅ)โˆ’1/16 log |๐‘ฅ+1/2+โˆš((๐‘ฅ)^2+๐‘ฅ)|+C_3 Putting the value of I_1 and I_2 in (1) , we get โˆซ1โ–’ใ€–๐‘ฅ โˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅ = 1/2 โˆซ1โ–’ใ€–(1+2๐‘ฅ) โˆš(๐‘ฅ+๐‘ฅ^2 )ใ€— ๐‘‘๐‘ฅโˆ’1/2 โˆซ1โ–’โˆš(๐‘ฅ+๐‘ฅ^2 ) ๐‘‘๐‘ฅ = ใ€–(๐‘ฅ+ ๐‘ฅ^2)ใ€—^(3/2)/3+C_1โˆ’((2๐‘ฅ + 1) โˆš(๐‘ฅ^2 + ๐‘ฅ))/8+1/16 logโก|๐‘ฅ+1/2+โˆš(๐‘ฅ^2+๐‘ฅ)|โˆ’C_3 = ใ€–(๐’™+ ๐’™^๐Ÿ)ใ€—^(๐Ÿ‘/๐Ÿ)/๐Ÿ‘โˆ’((๐Ÿ๐’™ + ๐Ÿ) โˆš(๐’™^๐Ÿ + ๐’™))/๐Ÿ–+๐Ÿ/๐Ÿ๐Ÿ” ๐’๐’๐’ˆโก|๐’™+๐Ÿ/๐Ÿ+โˆš(๐’™^๐Ÿ+๐’™)|+๐‚

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