Example 16 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 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
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo