Examples

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

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

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.