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Misc 14 - Integrate 1 / (x2 + 1) (x2 + 4) - Chapter 7 - Miscellaneous

Misc 14 - Chapter 7 Class 12 Integrals - Part 2
Misc 14 - Chapter 7 Class 12 Integrals - Part 3 Misc 14 - Chapter 7 Class 12 Integrals - Part 4 Misc 14 - Chapter 7 Class 12 Integrals - Part 5

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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 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.