Question 1 (Choice 1) - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Question 1 – Choice 1 Find ∫1▒〖log⁡𝑥/(1 + log⁡𝑥 )^2 𝑑𝑥〗Let 𝐈=∫1▒〖log⁡𝑥/(1 + log⁡𝑥 )^2 𝑑𝑥〗 =∫1▒〖(𝒍𝒐𝒈⁡𝒙 + 𝟏 − 𝟏)/(1 + log⁡𝑥 )^2 𝑑𝑥〗 =∫1▒〖((1 + log⁡𝑥) − 1)/(1 + log⁡𝑥 )^2 𝑑𝑥〗 =∫1▒〖((1 + log⁡𝑥) )/(1 + log⁡𝑥 )^2 𝑑𝑥〗−∫1▒〖1/(1 + log⁡𝑥 )^2 𝑑𝑥〗 =∫1▒〖(𝟏 )/((𝟏 + 𝒍𝒐𝒈⁡𝒙 ) ) 𝒅𝒙〗−∫1▒〖𝟏/(𝟏 + 𝒍𝒐𝒈⁡𝒙 )^𝟐 𝒅𝒙〗 Solving ∫1▒〖(𝟏 )/((𝟏 + 𝐥𝐨𝐠⁡𝐱 ) ) 𝐝𝐱〗 Using Integration by parts ∫1▒〖(𝟏 )/((𝟏 + 𝒍𝒐𝒈⁡𝒙 ) ) 𝒅𝒙〗 = ∫1▒〖(1 )/((1 + 𝑙𝑜𝑔⁡𝑥 ) ) × 1 𝑑𝑥〗 = 1/((1 + log⁡𝑥)) ∫1▒〖1 𝑑𝑥〗−∫1▒(𝒅(𝟏/(𝟏 + 𝒍𝒐𝒈 𝒙))/𝒅𝒙 ∫1▒〖𝟏 𝒅𝒙〗) 𝒅𝒙 = 1/((1 + log⁡𝑥))× 𝑥−∫1▒((−𝟏)/(𝟏 +𝒍𝒐𝒈 𝒙)^𝟐 ×𝟏/𝒙 × 𝒙) 𝒅𝒙 = 𝑥/((1 + log⁡𝑥))+∫1▒𝟏/(𝟏 + 𝒍𝒐𝒈 𝒙)^𝟐 𝒅𝒙We 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) = 1/(1 + log x) and g(x) = 1 Thus I=∫1▒〖(1 )/((1 + 𝑙𝑜𝑔⁡𝑥 ) ) 𝑑𝑥〗−∫1▒〖1/(1 + 𝑙𝑜𝑔⁡𝑥 )^2 𝑑𝑥〗 = 𝑥/((1 + log⁡𝑥))+∫1▒1/(1 + 𝑙𝑜𝑔 𝑥)^2 𝑑𝑥−∫1▒〖1/(1 + 𝑙𝑜𝑔⁡𝑥 )^2 𝑑𝑥〗 = 𝒙/((𝟏 + 𝒍𝒐𝒈⁡𝒙))+𝑪