Find ∫(x 3 + x  + 1)/((x 2 - 1) dx

 

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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