Check sibling questions

Ex 7.7, 13 (Supplementary NCERT) - Integrate x + 1 root (2x2 + 3)

Ex 7.7, 13 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 2
Ex 7.7, 13 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 3 Ex 7.7, 13 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 4 Ex 7.7, 13 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 5 Ex 7.7, 13 (Supplementary NCERT) - Chapter 7 Class 12 Integrals - Part 6

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Ex 7.7, 13 (Supplementary NCERT) ๐‘ฅ+1 โˆš(2๐‘ฅ^2+3) ๐‘ฅ+1 โˆš( 2๐‘ฅ^2+3) Integrating the function w.r.t.x โˆซ1โ–’ใ€–(๐‘ฅ+1) โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–[๐‘ฅโˆš(ใ€–2๐‘ฅใ€—^2+3)+1โˆš(ใ€–2๐‘ฅใ€—^2+3)] ๐‘‘๐‘ฅใ€— = โˆซ1โ–’ใ€–๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–โˆš(ใ€–2๐‘ฅใ€—^2+3 ) ๐‘‘๐‘ฅใ€—ใ€— Solving ๐ˆ_๐Ÿ ๐ˆ_๐Ÿ=โˆซ1โ–’ใ€–๐‘ฅ โˆš(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 = โˆซ1โ–’ใ€–๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ Putting the value of (ใ€–2๐‘ฅใ€—^2+3) and dx I_1 = โˆซ1โ–’ใ€–๐‘ฅ โˆš๐‘กใ€— . ๐‘‘๐‘ฅ I_1 = โˆซ1โ–’ใ€–๐‘ฅ โˆš๐‘กใ€— ๐‘‘๐‘ก/4๐‘ฅ I_1 = 1/4 โˆซ1โ–’ใ€– โˆš๐‘ก ใ€—. ๐‘‘๐‘ก I_1 = 1/4 โˆซ1โ–’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Ÿ I_2 = โˆซ1โ–’โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’โˆš(2 (ใ€– ๐‘ฅใ€—^2+3/2) ) ๐‘‘๐‘ฅ I_2 = โˆซ1โ–’ใ€–โˆš2 โˆš( ใ€– ๐‘ฅใ€—^2+3/2)ใ€— ๐‘‘๐‘ฅ I_2 =โˆš2 โˆซ1โ–’โˆš(ใ€– ๐‘ฅใ€—^2+3/2) ๐‘‘๐‘ฅ I_2 =โˆš2 โˆซ1โ–’โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 ) ๐‘‘๐‘ฅ It is of form โˆซ1โ–’ใ€–โˆš(๐‘ฅ^2+๐‘Ž^2 ) ใ€— ๐‘‘๐‘ฅ=1/2 ๐‘ฅ โˆš(๐‘ฅ^2+๐‘Ž^2 )+๐‘Ž^2/2 logโก|๐‘ฅ+โˆš(๐‘ฅ^2+๐‘Ž^2 )|+C_2 Replacing a by โˆš(3/2), we get I_2 =โˆš2 [๐‘ฅ/2 โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 )]+โˆš2 ใ€–((โˆš(3/2 )))/2ใ€—^2 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+(โˆš(3/2))^2 )|+C_2 I_2 =โˆš2 [๐‘ฅ/2 โˆš(ใ€– ๐‘ฅใ€—^2+3/2)]+โˆš2 (3/2)/2 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+C_2 I_2 =(โˆš2 ๐‘ฅ)/2.โˆš(ใ€– 2๐‘ฅใ€—^2+3)/โˆš2+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€–โˆš2 Cใ€—_2 I_2 =๐‘ฅ/2.โˆš(ใ€– 2๐‘ฅใ€—^2+3)+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€– Cใ€—_3 Now, put the value of I_1 and I_2 in eq. (1), we get โˆซ1โ–’ใ€–(๐‘ฅ+1) โˆš(ใ€–2๐‘ฅใ€—^2+3)ใ€— ๐‘‘๐‘ฅ=โˆซ1โ–’๐‘ฅ โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ+โˆซ1โ–’โˆš(ใ€–2๐‘ฅใ€—^2+3) ๐‘‘๐‘ฅ = (ใ€–2๐‘ฅใ€—^2+3)^(3/2)/6+ใ€– Cใ€—_1+๐‘ฅ/2 โˆš(ใ€–2๐‘ฅใ€—^2+3)+(3โˆš2)/4 logโก|๐‘ฅ+โˆš(ใ€– ๐‘ฅใ€—^2+3/2)|+ใ€– Cใ€—_3 = (ใ€–๐Ÿ๐’™ใ€—^๐Ÿ+๐Ÿ‘)^(๐Ÿ‘/๐Ÿ)/๐Ÿ”+๐’™/๐Ÿ โˆš(ใ€–๐Ÿ๐’™ใ€—^๐Ÿ+๐Ÿ‘)+(๐Ÿ‘โˆš๐Ÿ)/๐Ÿ’ ๐’๐’๐’ˆโก|๐’™+โˆš(ใ€– ๐’™ใ€—^๐Ÿ+๐Ÿ‘/๐Ÿ)|+๐‚

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.