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Chapter 7 Class 12 Integrals
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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

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Ex 7.6, 14 γ€–π‘₯(log⁑π‘₯)γ€—^2 ∫1β–’γ€–π‘₯(log⁑π‘₯ )^2.𝑑π‘₯ " " γ€— ∴ ∫1β–’γ€–π‘₯(log⁑π‘₯ )^2.𝑑π‘₯γ€—=∫1β–’γ€–(log⁑π‘₯ )^2 π‘₯ .𝑑π‘₯γ€— = (log⁑π‘₯ )^2 ∫1β–’γ€–π‘₯ .γ€— 𝑑π‘₯βˆ’βˆ«1β–’((𝑑(log⁑π‘₯ )^2)/𝑑π‘₯ ∫1β–’γ€–π‘₯ .𝑑π‘₯γ€—) 𝑑π‘₯ = (log⁑π‘₯ )^2 . π‘₯^2/2βˆ’βˆ«1β–’(2(log⁑π‘₯ ) 1/π‘₯ ∫1β–’γ€–π‘₯ .𝑑π‘₯γ€—) 𝑑π‘₯ Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓′(π‘₯)∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = x and g(x) = (log x)2 = π‘₯^2/2 (log⁑π‘₯ )^2βˆ’2∫1β–’γ€–log⁑π‘₯/π‘₯ . π‘₯^2/2γ€— 𝑑π‘₯ = π‘₯^2/2 (log⁑π‘₯ )^2βˆ’βˆ«1β–’γ€–π‘₯ log⁑π‘₯ γ€— 𝑑π‘₯ Solving I1 I1 = ∫1β–’γ€–π‘₯ log⁑π‘₯ γ€— 𝑑π‘₯ ∫1β–’γ€–π‘₯ log⁑π‘₯ γ€— 𝑑π‘₯=∫1β–’(log⁑π‘₯ )π‘₯ 𝑑π‘₯ =log⁑π‘₯ ∫1β–’π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’(𝑑(log⁑π‘₯ )/𝑑π‘₯ ∫1β–’γ€–π‘₯.𝑑π‘₯γ€—)𝑑π‘₯ Now we know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓′(π‘₯)∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = x and g(x) = log x =log⁑π‘₯ (π‘₯^2/2)βˆ’βˆ«1β–’γ€–1/π‘₯ . π‘₯^2/2. 𝑑π‘₯γ€— =γ€–π‘₯^2/2 log〗⁑〖 π‘₯γ€—βˆ’1/2 ∫1β–’γ€–π‘₯. 𝑑π‘₯γ€— =γ€–π‘₯^2/2 log〗⁑π‘₯βˆ’1/2 . π‘₯^2/2 +𝐢 =γ€–π‘₯^2/2 π‘™π‘œπ‘”γ€—β‘γ€– π‘₯γ€—βˆ’ π‘₯^2/4 +𝐢 Putting value of I1 in (1), ∫1β–’γ€–π‘₯(log⁑π‘₯ )^2.𝑑π‘₯γ€—=π‘₯^2/2 (log⁑π‘₯ )^2βˆ’βˆ«1β–’γ€– 𝒙 .π’π’π’ˆβ‘π’™ 𝒅𝒙〗 =π‘₯^2/2 (log⁑π‘₯ )^2βˆ’((π‘₯^2 (log⁑π‘₯ ))/2 βˆ’ π‘₯^2/4 +𝐢1) =π‘₯^2/2 (log⁑π‘₯ )^2βˆ’ (π‘₯^2 (log⁑π‘₯ ))/2 + π‘₯^2/4 βˆ’πΆ1 =𝒙^𝟐/𝟐 (π’π’π’ˆβ‘π’™ )^πŸβˆ’ (𝒙^𝟐 (π’π’π’ˆβ‘π’™ ))/𝟐 + 𝒙^𝟐/πŸ’+π‘ͺ " "

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.