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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 14 Integrate the function 1/((π‘₯^2 + 1)(π‘₯^2 + 4) ) Integrating 1/(π‘₯^2 + 1)(π‘₯^2 + 4) Let π‘₯^2=𝑦 1/(π‘₯^2 + 1)(π‘₯^2 + 4) =1/(𝑦 + 4)(𝑦 + 1) Hence we can write 1/(𝑦 + 4)(𝑦 + 1) =𝐴/((𝑦 + 4) )+𝐡/((𝑦 + 1) ) 1/(𝑦 + 4)(𝑦 + 1) =(𝐴(𝑦 + 1)+𝐡(𝑦 + 4))/(𝑦 + 4)(𝑦 + 1) Canceling denominators 1 = A(𝑦+1)+B(𝑦+4) Putting π’š=βˆ’πŸ 1= A (βˆ’1+1)+B(βˆ’1+4) 1= A Γ—0+ B (3) 1=3B B =1/3 Putting π’š=βˆ’πŸ’ 1= A(βˆ’4+1)+ B(βˆ’4+4) 1= A(βˆ’3)+ B(0) 1=βˆ’3A A =(βˆ’1)/3 Therefore we can write 1/(𝑦 + 4)(𝑦 + 1) =𝐴/((𝑦 + 4) )+𝐡/((𝑦 + 1) ) 1/(𝑦 + 4)(𝑦 + 1) =(((βˆ’1)/( 3)))/((𝑦 + 4) )+((1/3))/((𝑦 + 1) ) Putting back 𝑦=π‘₯^2 1/(π‘₯^2 + 4)(π‘₯^2 + 1) =(((βˆ’1)/( 3)))/((π‘₯^2 + 4) )+((( 1)/( 3)))/((π‘₯^2+ 1) ) = (βˆ’1)/3(π‘₯^2 + 4) +1/3(π‘₯^(2 )+ 1) Integrating w.r.t. π‘₯ ∫1β–’β–ˆ(1/(π‘₯^(2 )+ 4)(π‘₯^2 + 1) 𝑑π‘₯) =∫1β–’((βˆ’1)/3(π‘₯^2 + 4) +1/3(π‘₯^2 + 1) )𝑑π‘₯ =∫1β–’γ€–(βˆ’1)/3(π‘₯^2 + 4) 𝑑π‘₯+γ€— ∫1β–’γ€–(βˆ’1)/3(π‘₯^2 + 1) 𝑑π‘₯γ€— =(βˆ’1)/( 3) ∫1β–’γ€–1/(π‘₯^2 + 2^2 ) 𝑑π‘₯+1/3 ∫1β–’γ€–1/(π‘₯^2 + 1^2 ) 𝑑π‘₯γ€—γ€— =(βˆ’1)/3 Γ— 1/2 tan^(βˆ’1)⁑〖π‘₯/2+1/3 Γ— 1/1 tan^(βˆ’1)⁑〖π‘₯/1+𝐢〗 γ€— =(βˆ’1)/( 6) tan^(βˆ’1)⁑〖π‘₯/2+1/3 tan^(βˆ’1)⁑〖π‘₯+𝑐〗 γ€— We know that ∫1β–’γ€–1/(π‘₯^2 + π‘Ž^2 ) 𝑑π‘₯=1/π‘Ž tan^(βˆ’1)⁑〖π‘₯/π‘Ž+𝑐〗 γ€— =𝟏/πŸ‘ γ€–π­πšπ§γ€—^(βˆ’πŸ)β‘γ€–π’™βˆ’πŸ/πŸ” γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑〖𝒙/𝟐+𝒄〗 γ€—

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.