Find ∫(x 3 + x  + 1)/((x 2 - 1) dx

 

[Integration Class 12] Find: ∫ (x^3 + x + 1) / (x^2 - 1) dx - Teachoo - CBSE Class 12 Sample Paper for 2023 Boards

part 2 - Question 31 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 31 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 4 - Question 31 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 5 - Question 31 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

 

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Question 31 Find ∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗 = ∫1▒〖𝒙+(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝒅𝒙〗 = ∫1▒〖𝑥 𝑑𝑥〗+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝑑𝑥〗 = 𝑥^2/2+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙^𝟐 − 𝟏)) 𝑑𝑥〗 = 𝒙^𝟐/𝟐+∫1▒〖(𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏)) 𝒅𝒙〗 Now Solving (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 − 𝟏)) (2𝑥 + 1)/((𝑥 + 1)(𝑥 − 1) ) = (𝐴(𝑥 − 1) + 𝐵(𝑥 + 1))/((𝑥 + 1)(𝑥 − 1) ) Cancelling denominator 2𝑥+1=𝐴(𝑥−1)+𝐵(𝑥+1) Putting x = 1 in (2) 2𝑥+1=𝐴(𝑥−1)+𝐵(𝑥+1) 2(1)+1 = 𝐴(1−1) + 𝐵(1+1) 3 = A × 0+2𝐵 3 = 2𝐵 𝑩=𝟑/𝟐 Putting x = −1 in (2) 2𝑥+1=𝐴(=1−1)+𝐵(𝑥+1) 2(−1)+1 = 𝐴(−1−1) + 𝐵(−1+1) −2+1 = A × −2+𝐵 × 0 −1 = −2A 1 = 2A 1/2 = A 𝑨=𝟏/𝟐 Hence we can write it as (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 − 𝟏)) (𝟐𝒙 + 𝟏)/((𝒙 + 𝟏)(𝒙 − 𝟏) ) = 𝟏/(𝟐(𝒙 + 𝟏)) + 𝟑/(𝟐(𝒙 − 𝟏)) Therefore , from (1) we get, ∫1▒〖((𝑥^3 + 𝑥 + 1))/((𝑥^2 − 1)) 𝑑𝑥〗 =𝑥^2/2+ ∫1▒(1/(2(𝑥 + 1)) " + " 3/(2(𝑥 − 1))) 𝑑𝑥 =𝑥^2/2+ ∫1▒𝑑𝑥/(2(𝑥 + 1))+∫1▒3𝑑𝑥/(2(𝑥 − 1)) =𝒙^𝟐/𝟐+𝟏/𝟐 ∫1▒〖𝒅𝒙/((𝒙 + 𝟏)) + 𝟑/𝟐〗 ∫1▒𝒅𝒙/((𝒙 − 𝟏)) =𝑥^2/2 +1/2 log⁡|(𝑥+1)|+3/2 log⁡|𝑥−1|+𝐶 =𝑥^2/2+1/2 ( log⁡|(𝑥+1)|+3 log⁡|𝑥−1| )+𝐶 =𝑥^2/2+1/2 ( log⁡|(𝑥+1) (𝑥−1)^3 | )+𝐶

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo