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


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Example 37 Evaluate ∫1β–’(π‘₯^4 𝑑π‘₯)/(π‘₯ βˆ’1)(π‘₯^2 + 1) Let I = ∫1β–’(π‘₯^4 𝑑π‘₯)/(π‘₯ βˆ’1)(π‘₯^2 + 1) 𝑑π‘₯ We can write π‘₯^4/(π‘₯ βˆ’1)(π‘₯^2 + 1) = π‘₯^4/(π‘₯^3 βˆ’ π‘₯^2+ π‘₯ βˆ’ 1) Dividing Numerator by denominator as follows. Hence π‘₯^4 = (π‘₯^3βˆ’π‘₯^2+π‘₯+1) (π‘₯+1)+1 Thus, π‘₯^4/(π‘₯^3 βˆ’ π‘₯^2 + π‘₯ + 1) = (π‘₯+1)+1/(π‘₯^3 βˆ’ π‘₯^2 + π‘₯ + 1) = (π‘₯+1)+1/((π‘₯ βˆ’ 1) (π‘₯^2 +1) ) Now, we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= ((𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1))/((π‘₯^2 + 1)(π‘₯ βˆ’1)) Canceling denominator 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) Putting x = 1 1 = (𝐴(1) + 𝐡)(1βˆ’1) + 𝐢 ((βˆ’1)^2 + 1) 1 = (𝐴+𝐡)(0)+ 𝐢 (1+1) 1 = 2𝐢 𝐢=1/2 Putting x = 0 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) 1 = (𝐴(0) + 𝐡)(0βˆ’1) + 𝐢 (0^2+1) 1 = (𝐡)(βˆ’1) + 𝐢 (1) 1 = 𝐢 βˆ’"B" B =πΆβˆ’1 B =1/2 βˆ’1 B =(βˆ’1)/2 Putting x = βˆ’ 1 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) 1 = (𝐴(βˆ’1)+ 𝐡)(βˆ’1βˆ’1) + 𝐢 ((βˆ’1)^2+1) 1 = (βˆ’π΄+𝐡)(βˆ’2)+𝐢 (1+1) 1 = (π΄βˆ’π΅)2+𝐢 (2) 1/2=π΄βˆ’π΅+𝐢 𝐴=1/2+π΅βˆ’πΆ 𝐴 =1/2βˆ’1/2βˆ’1/2 𝐴 =(βˆ’1)/2 Hence we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) ) = (βˆ’ 1/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) + (1/2)/(π‘₯ βˆ’ 1) = (βˆ’1)/2 ( π‘₯)/(π‘₯^2 + 1) βˆ’1/2 1/(π‘₯^2 + 1)+ 1/2(π‘₯ βˆ’ 1) Hence we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) ) = (βˆ’ 1/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) + (1/2)/(π‘₯ βˆ’ 1) = (βˆ’1)/2 ( π‘₯)/(π‘₯^2 + 1) βˆ’1/2 1/(π‘₯^2 + 1)+ 1/2(π‘₯ βˆ’ 1) Therefore, we can write I=∫1β–’γ€–(π‘₯+1)+1/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) 𝑑π‘₯γ€— =∫1β–’[(π‘₯+1)βˆ’1/2 π‘₯/((π‘₯^2 + 1) ) 𝑑π‘₯βˆ’βˆ«1β–’γ€–1/2 1/(π‘₯^2 + 1) 𝑑π‘₯+∫1β–’γ€–1/2 1/((π‘₯ βˆ’ 1) ) 𝑑π‘₯γ€—γ€—] =π‘₯^2/2+π‘₯βˆ’1/2 ∫1β–’γ€–π‘₯/(π‘₯^2 + 1)βˆ’1/2 ∫1β–’γ€–1/(π‘₯^2 + 1) 𝑑π‘₯+1/2 ∫1β–’γ€–1/(π‘₯ βˆ’ 1) 𝑑π‘₯γ€—γ€—γ€— ∴ I = π‘₯^2/2+π‘₯ – 1/2 I"1 βˆ’ " 1/2 I"2 + " 1/2 I"3" Solving π‘°πŸ I1=∫1β–’γ€–π‘₯/(π‘₯^2 + 1) 𝑑π‘₯γ€— Put 𝑑=π‘₯^2+1 Differentiating w.r.t. π‘₯ 𝑑𝑑/𝑑π‘₯=2π‘₯+0 𝑑𝑑/2π‘₯=𝑑π‘₯ Therefore, ∫1β–’γ€–(π‘₯ 𝑑π‘₯)/(π‘₯^2 + 1)=∫1β–’π‘₯/𝑑 𝑑𝑑/2π‘₯γ€—=∫1β–’1/2 𝑑𝑑/𝑑=1/2 π‘™π‘œπ‘”|𝑑|+𝐢1 Putting 𝑑=π‘₯^2+1 =1/2 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1 And, I2=∫1β–’γ€–1/(π‘₯^2 + 1) 𝑑π‘₯γ€—=tan^(βˆ’1)⁑〖π‘₯+γ€— 𝐢2 I3=∫1β–’γ€–1/(π‘₯ βˆ’1) 𝑑π‘₯γ€—=π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3 Hence 𝐼=π‘₯^2/2+π‘₯βˆ’1/2 𝐼1βˆ’1/2 𝐼2+1/2 𝐼3 =π‘₯^2/2+π‘₯βˆ’1/2 (1/2 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1)βˆ’1/2 (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (π‘₯)+C_2 )βˆ’1/2 (π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3) =π‘₯^2/2+π‘₯βˆ’1/4 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1/2βˆ’1/2 tan^(βˆ’1)⁑〖π‘₯ 𝐢2/2+1/2 π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3/2γ€— =𝒙^𝟐/𝟐+𝒙+𝟏/𝟐 π’π’π’ˆ|π’™βˆ’πŸ|βˆ’πŸ/πŸ’ π’π’π’ˆ(𝒙^𝟐+𝟏)βˆ’πŸ/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖𝒙+π‘ͺγ€—

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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.