Ex 7.2, 38 - Integration 10x9 + 10x loge 10 / x10 + 10x - Ex 7.2

Ex 7.2, 38 - Chapter 7 Class 12 Integrals - Part 2

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Ex 7.2, 38 ∫1β–’(γ€–10π‘₯γ€—^9+γ€–10γ€—^π‘₯ log_𝑒⁑10)/(π‘₯10+ 10π‘₯) dx equals (A) 10π‘₯ – π‘₯^10 + 𝐢 (B) 10π‘₯+π‘₯^10+𝐢 (C) (10π‘₯ – π‘₯^10 )^(βˆ’1) + 𝐢 (D) log⁑(10π‘₯+π‘₯10) + 𝐢 Let π‘₯10+ 10π‘₯= 𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ γ€–10π‘₯γ€—^(10βˆ’1)+γ€–10γ€—^π‘₯ π‘™π‘œπ‘”β‘10= 𝑑𝑑/𝑑π‘₯ γ€–10π‘₯γ€—^9+γ€–10γ€—^π‘₯ π‘™π‘œπ‘”β‘10= 𝑑𝑑/𝑑π‘₯ 𝑑π‘₯= 𝑑𝑑/(γ€–10π‘₯γ€—^9 + γ€–10γ€—^π‘₯ π‘™π‘œπ‘”β‘10 ) (Using (π‘Ž^π‘₯ )^β€²=π‘Ž^π‘₯ π‘™π‘œπ‘”β‘π‘Ž) Now, our function becomes ∫1β–’γ€–" " (10π‘₯γ€–9+10γ€—^π‘₯ π‘™π‘œπ‘”β‘10)/(π‘₯^10 + γ€–10γ€—^π‘₯ )γ€— . 𝑑π‘₯ Putting (π‘₯^10+ γ€–10γ€—^π‘₯ )=𝑑 & 𝑑π‘₯=" " 𝑑𝑑/(γ€–10π‘₯γ€—^9 + γ€–10γ€—^π‘₯ π‘™π‘œπ‘”β‘10 ) = ∫1β–’γ€–" " (10π‘₯γ€–9+10γ€—^π‘₯ π‘™π‘œπ‘”β‘10)/𝑑〗 . 𝑑𝑑/(γ€–10π‘₯γ€—^9+γ€–10γ€—^π‘₯ π‘™π‘œπ‘”β‘10 ) " " = ∫1β–’γ€–" " 1/𝑑〗.𝑑𝑑 = log |𝑑|+𝐢 = log |γ€–10γ€—^π‘₯+ π‘₯^10 |+𝐢 = log (γ€–10γ€—^π‘₯+ π‘₯^10 )+𝐢 ∴ Option D is correct. (Using 𝑑=γ€–10γ€—^π‘₯+π‘₯^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 10 years. He provides courses for Maths and Science at Teachoo.