Find ∫(x ^{ 3 } + x + 1)/((x ^{ 2 } - 1) dx
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CBSE Class 12 Sample Paper for 2023 Boards
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CBSE Class 12 Sample Paper for 2023 Boards
Last updated at March 28, 2023 by Teachoo
Get live Maths 1-on-1 Classs - Class 6 to 12
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