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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Ex 7.2, 38 โˆซ1โ–’(ใ€–10๐‘ฅใ€—^9+ใ€–10ใ€—^๐‘ฅ log_๐‘’โก10)/(๐‘ฅ10+ 10๐‘ฅ) dx equals (A) 10๐‘ฅ โ€“ ๐‘ฅ^10 + ๐ถ (B) 10๐‘ฅ+๐‘ฅ^10+๐ถ (C) (10๐‘ฅ โ€“ ๐‘ฅ^10 )^(โˆ’1) + ๐ถ (D) logโก(10๐‘ฅ+๐‘ฅ10) + ๐ถ Let ๐‘ฅ10+ 10๐‘ฅ= ๐‘ก Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ใ€–10๐‘ฅใ€—^(10โˆ’1)+ใ€–10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ใ€–10๐‘ฅใ€—^9+ใ€–10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10= ๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ= ๐‘‘๐‘ก/(ใ€–10๐‘ฅใ€—^9 + ใ€–10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10 ) (Using (๐‘Ž^๐‘ฅ )^โ€ฒ=๐‘Ž^๐‘ฅ ๐‘™๐‘œ๐‘”โก๐‘Ž) Now, our function becomes โˆซ1โ–’ใ€–" " (10๐‘ฅใ€–9+10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10)/(๐‘ฅ^10 + ใ€–10ใ€—^๐‘ฅ )ใ€— . ๐‘‘๐‘ฅ Putting (๐‘ฅ^10+ ใ€–10ใ€—^๐‘ฅ )=๐‘ก & ๐‘‘๐‘ฅ=" " ๐‘‘๐‘ก/(ใ€–10๐‘ฅใ€—^9 + ใ€–10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10 ) = โˆซ1โ–’ใ€–" " (10๐‘ฅใ€–9+10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10)/๐‘กใ€— . ๐‘‘๐‘ก/(ใ€–10๐‘ฅใ€—^9+ใ€–10ใ€—^๐‘ฅ ๐‘™๐‘œ๐‘”โก10 ) " " = โˆซ1โ–’ใ€–" " 1/๐‘กใ€—.๐‘‘๐‘ก = log |๐‘ก|+๐ถ = log |ใ€–10ใ€—^๐‘ฅ+ ๐‘ฅ^10 |+๐ถ = log (ใ€–10ใ€—^๐‘ฅ+ ๐‘ฅ^10 )+๐ถ โˆด Option D is correct. (Using ๐‘ก=ใ€–10ใ€—^๐‘ฅ+๐‘ฅ^10) (As 10x and x10 are positive)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.