Check sibling questions

Find ∫(x 3 + x  + 1)/((x 2 - 1) dx

 

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Question 31 Find ∫1β–’γ€–((π‘₯^3 + π‘₯ + 1))/((π‘₯^2 βˆ’ 1)) 𝑑π‘₯γ€—βˆ«1β–’γ€–((π‘₯^3 + π‘₯ + 1))/((π‘₯^2 βˆ’ 1)) 𝑑π‘₯γ€— = ∫1▒〖𝒙+(πŸπ’™ + 𝟏)/((𝒙^𝟐 βˆ’ 𝟏)) 𝒅𝒙〗 = ∫1β–’γ€–π‘₯ 𝑑π‘₯γ€—+∫1β–’γ€–(πŸπ’™ + 𝟏)/((𝒙^𝟐 βˆ’ 𝟏)) 𝑑π‘₯γ€— = π‘₯^2/2+∫1β–’γ€–(πŸπ’™ + 𝟏)/((𝒙^𝟐 βˆ’ 𝟏)) 𝑑π‘₯γ€— = 𝒙^𝟐/𝟐+∫1β–’γ€–(πŸπ’™ + 𝟏)/((𝒙 + 𝟏)(𝒙 βˆ’ 𝟏)) 𝒅𝒙〗 Now Solving (πŸπ’™ + 𝟏)/((𝒙 + 𝟏)(𝒙 βˆ’ 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 βˆ’ 𝟏)) (2π‘₯ + 1)/((π‘₯ + 1)(π‘₯ βˆ’ 1) ) = (𝐴(π‘₯ βˆ’ 1) + 𝐡(π‘₯ + 1))/((π‘₯ + 1)(π‘₯ βˆ’ 1) ) Cancelling denominator 2π‘₯+1=𝐴(π‘₯βˆ’1)+𝐡(π‘₯+1) Putting x = 1 in (2) 2π‘₯+1=𝐴(π‘₯βˆ’1)+𝐡(π‘₯+1) 2(1)+1 = 𝐴(1βˆ’1) + 𝐡(1+1) 3 = A Γ— 0+2𝐡 3 = 2𝐡 𝑩=πŸ‘/𝟐 Putting x = βˆ’1 in (2) 2π‘₯+1=𝐴(=1βˆ’1)+𝐡(π‘₯+1) 2(βˆ’1)+1 = 𝐴(βˆ’1βˆ’1) + 𝐡(βˆ’1+1) βˆ’2+1 = A Γ— βˆ’2+𝐡 Γ— 0 βˆ’1 = βˆ’2A 1 = 2A 1/2 = A 𝑨=𝟏/𝟐 Hence we can write it as (πŸπ’™ + 𝟏)/((𝒙 + 𝟏)(𝒙 βˆ’ 𝟏) ) = 𝑨/((𝒙 + 𝟏)) + 𝑩/((𝒙 βˆ’ 𝟏)) (πŸπ’™ + 𝟏)/((𝒙 + 𝟏)(𝒙 βˆ’ 𝟏) ) = 𝟏/(𝟐(𝒙 + 𝟏)) + πŸ‘/(𝟐(𝒙 βˆ’ 𝟏)) Therefore , from (1) we get, ∫1β–’γ€–((π‘₯^3 + π‘₯ + 1))/((π‘₯^2 βˆ’ 1)) 𝑑π‘₯γ€— =π‘₯^2/2+ ∫1β–’(1/(2(π‘₯ + 1)) " + " 3/(2(π‘₯ βˆ’ 1))) 𝑑π‘₯ =π‘₯^2/2+ ∫1▒𝑑π‘₯/(2(π‘₯ + 1))+∫1β–’3𝑑π‘₯/(2(π‘₯ βˆ’ 1)) =𝒙^𝟐/𝟐+𝟏/𝟐 ∫1▒〖𝒅𝒙/((𝒙 + 𝟏)) + πŸ‘/πŸγ€— ∫1▒𝒅𝒙/((𝒙 βˆ’ 𝟏)) =π‘₯^2/2 +1/2 log⁑|(π‘₯+1)|+3/2 log⁑|π‘₯βˆ’1|+𝐢 =π‘₯^2/2+1/2 ( log⁑|(π‘₯+1)|+3 log⁑|π‘₯βˆ’1| )+𝐢 =π‘₯^2/2+1/2 ( log⁑|(π‘₯+1) (π‘₯βˆ’1)^3 | )+𝐢

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.