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Integration by partial fraction - Type 5
Integration by partial fraction - Type 5
Last updated at Dec. 20, 2019 by Teachoo
Example 16 Find ∫1▒(𝑥^2+ 𝑥 +1 𝑑𝑥 )/((𝑥 + 2) (𝑥^2+1) ) We can write equation as (𝑥^2+ 𝑥 + 1)/((𝑥 + 1) (𝑥 + 2) )=𝐴/(𝑥 + 2) + (𝐵𝑥 + 𝐶)/(𝑥^2+ 1) Cancelling denominator 〖 𝑥〗^2+ 𝑥+1=𝐴(𝑥^2+1)+(𝐵𝑥+𝐶) (𝑥+2) Putting x = −𝟐 (−2)^2+(−2)+1=𝐴((−2)^2+1)+0 4−2+1= 5A 3/5 = A Putting x = 𝟎 𝑥^2+ 𝑥+1=𝐴(𝑥^2+1)+(𝐵𝑥+𝐶) (𝑥+2) 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 3/5 + 2C 1 – 3/5 = 2C 2/5 = 2C C = 1/5 Putting x = 1 𝑥^2+ 𝑥+1=𝐴(𝑥^2+1)+(𝐵𝑥+𝐶) (𝑥+2) 1+1+1= 2A + (B + C)(3) 3 = 2A + 3 (B + C) 3 = 2(3/5) + 3 (B+1/5) 3 – 6/5 = 3 (B+1/5) 9/5 = 3 (B+1/5) 3/5 – 1/5 = B B = 2/5 Thus, (𝑥^2+ 𝑥 + 1)/((𝑥 + 1) (𝑥 + 2) )=𝐴/(𝑥 + 2) + (𝐵𝑥 + 𝐶)/(𝑥^2+ 1) (𝑥^2+ 𝑥 + 1)/((𝑥 + 1)(𝑥^2+ 1)) = 3/(5 (𝑥 + 2)) + (1 (2𝑥 + 1))/(5 (𝑥^2 + 1)) Hence, our equation becomes ∫1▒(𝑥^2+ 𝑥 + 1)/((𝑥 + 2) (𝑥^2 + 1)) 𝑑𝑥= ∫1▒3/(5(𝑥^2 + 1)) 𝑑𝑥+∫1▒1/5 ((2𝑥 + 1))/(𝑥^2 + 1) 𝑑𝑥 = ∫1▒3/(5(𝑥^2 + 1)) 𝑑𝑥+ 1/5 ∫1▒〖2𝑥/(𝑥^2 + 1) 𝑑𝑥+〗 1/5 ∫1▒1/(𝑥^2 + 1) 𝑑𝑥 𝐈𝟐 1/5 ∫1▒2𝑥/(𝑥^2+ 1) 𝑑𝑥 Let 𝑡=𝑥^2+ 1 𝑑𝑡/𝑑𝑥=2𝑥 𝑑𝑡=2𝑥 𝑑𝑥 Substituting, =1/5 ∫1▒𝑑𝑡/𝑡 = 1/5 log |𝑡| + C_2 = 1/5 log |𝑥^2+1| + C_2 𝐈𝟑 1/5 ∫1▒1/(𝑥^2+ 1) 𝑑𝑥 = 1/5 〖𝑡𝑎𝑛〗^(−1) (𝑥)+C_3 Hence ∫1▒(𝑥^2+ 𝑥 + 1)/((𝑥 + 2) (𝑥^2+ 1)) 𝑑𝑥 =𝟑/𝟓 𝒍𝒐𝒈|𝒙+𝟐|+𝟏/𝟓 𝒍𝒐𝒈|𝒙^𝟐+𝟏|+𝟏/𝟓 〖𝒕𝒂𝒏〗^(−𝟏) (𝒙)+ C where C = C_1+ C_2+C_3