# Ex 7.6, 14 - Chapter 7 Class 12 Integrals

Last updated at April 16, 2024 by Teachoo

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Ex 7.6, 14 Important You are here

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Ex 7.6, 23 (MCQ)

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Last updated at April 16, 2024 by Teachoo

Ex 7.6, 14 〖𝑥(log𝑥)〗^2 ∫1▒〖𝑥(log𝑥 )^2.𝑑𝑥 " " 〗 ∴ ∫1▒〖𝑥(log𝑥 )^2.𝑑𝑥〗=∫1▒〖(log𝑥 )^2 𝑥 .𝑑𝑥〗 = (log𝑥 )^2 ∫1▒〖𝑥 .〗 𝑑𝑥−∫1▒((𝑑(log𝑥 )^2)/𝑑𝑥 ∫1▒〖𝑥 .𝑑𝑥〗) 𝑑𝑥 = (log𝑥 )^2 . 𝑥^2/2−∫1▒(2(log𝑥 ) 1/𝑥 ∫1▒〖𝑥 .𝑑𝑥〗) 𝑑𝑥 Now we know that ∫1▒〖𝑓(𝑥) 𝑔(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓′(𝑥)∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = x and g(x) = (log x)2 = 𝑥^2/2 (log𝑥 )^2−2∫1▒〖log𝑥/𝑥 . 𝑥^2/2〗 𝑑𝑥 = 𝑥^2/2 (log𝑥 )^2−∫1▒〖𝑥 log𝑥 〗 𝑑𝑥 Solving I1 I1 = ∫1▒〖𝑥 log𝑥 〗 𝑑𝑥 ∫1▒〖𝑥 log𝑥 〗 𝑑𝑥=∫1▒(log𝑥 )𝑥 𝑑𝑥 =log𝑥 ∫1▒𝑥 𝑑𝑥−∫1▒(𝑑(log𝑥 )/𝑑𝑥 ∫1▒〖𝑥.𝑑𝑥〗)𝑑𝑥 Now we know that ∫1▒〖𝑓(𝑥) 𝑔(𝑥) 〗 𝑑𝑥=𝑓(𝑥) ∫1▒𝑔(𝑥) 𝑑𝑥−∫1▒(𝑓′(𝑥)∫1▒𝑔(𝑥) 𝑑𝑥) 𝑑𝑥 Putting f(x) = x and g(x) = log x =log𝑥 (𝑥^2/2)−∫1▒〖1/𝑥 . 𝑥^2/2. 𝑑𝑥〗 =〖𝑥^2/2 log〗〖 𝑥〗−1/2 ∫1▒〖𝑥. 𝑑𝑥〗 =〖𝑥^2/2 log〗𝑥−1/2 . 𝑥^2/2 +𝐶 =〖𝑥^2/2 𝑙𝑜𝑔〗〖 𝑥〗− 𝑥^2/4 +𝐶 Putting value of I1 in (1), ∫1▒〖𝑥(log𝑥 )^2.𝑑𝑥〗=𝑥^2/2 (log𝑥 )^2−∫1▒〖 𝒙 .𝒍𝒐𝒈𝒙 𝒅𝒙〗 =𝑥^2/2 (log𝑥 )^2−((𝑥^2 (log𝑥 ))/2 − 𝑥^2/4 +𝐶1) =𝑥^2/2 (log𝑥 )^2− (𝑥^2 (log𝑥 ))/2 + 𝑥^2/4 −𝐶1 =𝒙^𝟐/𝟐 (𝒍𝒐𝒈𝒙 )^𝟐− (𝒙^𝟐 (𝒍𝒐𝒈𝒙 ))/𝟐 + 𝒙^𝟐/𝟒+𝑪 " "