Check sibling questions

Ex 7.6, 14 Class 12 Maths - NCERT Solutions - Integrate x (log x)^2

Ex 7.6, 14 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.6, 14 - Chapter 7 Class 12 Integrals - Part 3

Learn Intergation from Davneet Sir - Live lectures starting soon!


Transcript

Ex 7.6, 14 ใ€–๐‘ฅ(logโก๐‘ฅ)ใ€—^2 โˆซ1โ–’ใ€–๐‘ฅ(logโก๐‘ฅ )^2.๐‘‘๐‘ฅ " " ใ€— โˆด โˆซ1โ–’ใ€–๐‘ฅ(logโก๐‘ฅ )^2.๐‘‘๐‘ฅใ€—=โˆซ1โ–’ใ€–(logโก๐‘ฅ )^2 ๐‘ฅ .๐‘‘๐‘ฅใ€— = (logโก๐‘ฅ )^2 โˆซ1โ–’ใ€–๐‘ฅ .ใ€— ๐‘‘๐‘ฅโˆ’โˆซ1โ–’((๐‘‘(logโก๐‘ฅ )^2)/๐‘‘๐‘ฅ โˆซ1โ–’ใ€–๐‘ฅ .๐‘‘๐‘ฅใ€—) ๐‘‘๐‘ฅ = (logโก๐‘ฅ )^2 . ๐‘ฅ^2/2โˆ’โˆซ1โ–’(2(logโก๐‘ฅ ) 1/๐‘ฅ โˆซ1โ–’ใ€–๐‘ฅ .๐‘‘๐‘ฅใ€—) ๐‘‘๐‘ฅ Now we know that โˆซ1โ–’ใ€–๐‘“(๐‘ฅ) ๐‘”โก(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=๐‘“(๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(๐‘“โ€ฒ(๐‘ฅ)โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ) ๐‘‘๐‘ฅ Putting f(x) = x and g(x) = (log x)2 = ๐‘ฅ^2/2 (logโก๐‘ฅ )^2โˆ’2โˆซ1โ–’ใ€–logโก๐‘ฅ/๐‘ฅ . ๐‘ฅ^2/2ใ€— ๐‘‘๐‘ฅ = ๐‘ฅ^2/2 (logโก๐‘ฅ )^2โˆ’โˆซ1โ–’ใ€–๐‘ฅ logโก๐‘ฅ ใ€— ๐‘‘๐‘ฅ Solving I1 I1 = โˆซ1โ–’ใ€–๐‘ฅ logโก๐‘ฅ ใ€— ๐‘‘๐‘ฅ โˆซ1โ–’ใ€–๐‘ฅ logโก๐‘ฅ ใ€— ๐‘‘๐‘ฅ=โˆซ1โ–’(logโก๐‘ฅ )๐‘ฅ ๐‘‘๐‘ฅ =logโก๐‘ฅ โˆซ1โ–’๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(๐‘‘(logโก๐‘ฅ )/๐‘‘๐‘ฅ โˆซ1โ–’ใ€–๐‘ฅ.๐‘‘๐‘ฅใ€—)๐‘‘๐‘ฅ Now we know that โˆซ1โ–’ใ€–๐‘“(๐‘ฅ) ๐‘”โก(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=๐‘“(๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(๐‘“โ€ฒ(๐‘ฅ)โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ) ๐‘‘๐‘ฅ Putting f(x) = x and g(x) = log x =logโก๐‘ฅ (๐‘ฅ^2/2)โˆ’โˆซ1โ–’ใ€–1/๐‘ฅ . ๐‘ฅ^2/2. ๐‘‘๐‘ฅใ€— =ใ€–๐‘ฅ^2/2 logใ€—โกใ€– ๐‘ฅใ€—โˆ’1/2 โˆซ1โ–’ใ€–๐‘ฅ. ๐‘‘๐‘ฅใ€— =ใ€–๐‘ฅ^2/2 logใ€—โก๐‘ฅโˆ’1/2 . ๐‘ฅ^2/2 +๐ถ =ใ€–๐‘ฅ^2/2 ๐‘™๐‘œ๐‘”ใ€—โกใ€– ๐‘ฅใ€—โˆ’ ๐‘ฅ^2/4 +๐ถ Putting value of I1 in (1), โˆซ1โ–’ใ€–๐‘ฅ(logโก๐‘ฅ )^2.๐‘‘๐‘ฅใ€—=๐‘ฅ^2/2 (logโก๐‘ฅ )^2โˆ’โˆซ1โ–’ใ€– ๐’™ .๐’๐’๐’ˆโก๐’™ ๐’…๐’™ใ€— =๐‘ฅ^2/2 (logโก๐‘ฅ )^2โˆ’((๐‘ฅ^2 (logโก๐‘ฅ ))/2 โˆ’ ๐‘ฅ^2/4 +๐ถ1) =๐‘ฅ^2/2 (logโก๐‘ฅ )^2โˆ’ (๐‘ฅ^2 (logโก๐‘ฅ ))/2 + ๐‘ฅ^2/4 โˆ’๐ถ1 =๐’™^๐Ÿ/๐Ÿ (๐’๐’๐’ˆโก๐’™ )^๐Ÿโˆ’ (๐’™^๐Ÿ (๐’๐’๐’ˆโก๐’™ ))/๐Ÿ + ๐’™^๐Ÿ/๐Ÿ’+๐‘ช " "

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.