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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

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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 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.