Check sibling questions

Example 16 - Find integral x2 + x + 1 dx / (x + 2) (x2 + 1)

Example 16 - Chapter 7 Class 12 Integrals - Part 2
Example 16 - Chapter 7 Class 12 Integrals - Part 3
Example 16 - Chapter 7 Class 12 Integrals - Part 4
Example 16 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Example 16 Find ∫1β–’(π‘₯^2+ π‘₯ +1 𝑑π‘₯ )/((π‘₯ + 2) (π‘₯^2+1) ) We can write equation as (π‘₯^2+ π‘₯ + 1)/((π‘₯ + 1) (π‘₯ + 2) )=𝐴/(π‘₯ + 2) + (𝐡π‘₯ + 𝐢)/(π‘₯^2+ 1) Cancelling denominator γ€– π‘₯γ€—^2+ π‘₯+1=𝐴(π‘₯^2+1)+(𝐡π‘₯+𝐢) (π‘₯+2) Putting x = βˆ’πŸ (βˆ’2)^2+(βˆ’2)+1=𝐴((βˆ’2)^2+1)+0 4βˆ’2+1= 5A 3/5 = A Putting x = 𝟎 π‘₯^2+ π‘₯+1=𝐴(π‘₯^2+1)+(𝐡π‘₯+𝐢) (π‘₯+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 π‘₯^2+ π‘₯+1=𝐴(π‘₯^2+1)+(𝐡π‘₯+𝐢) (π‘₯+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, (π‘₯^2+ π‘₯ + 1)/((π‘₯ + 1) (π‘₯ + 2) )=𝐴/(π‘₯ + 2) + (𝐡π‘₯ + 𝐢)/(π‘₯^2+ 1) (π‘₯^2+ π‘₯ + 1)/((π‘₯ + 1)(π‘₯^2+ 1)) = 3/(5 (π‘₯ + 2)) + (1 (2π‘₯ + 1))/(5 (π‘₯^2 + 1)) Hence, our equation becomes ∫1β–’(π‘₯^2+ π‘₯ + 1)/((π‘₯ + 2) (π‘₯^2 + 1)) 𝑑π‘₯= ∫1β–’3/(5(π‘₯^2 + 1)) 𝑑π‘₯+∫1β–’1/5 ((2π‘₯ + 1))/(π‘₯^2 + 1) 𝑑π‘₯ = ∫1β–’3/(5(π‘₯^2 + 1)) 𝑑π‘₯+ 1/5 ∫1β–’γ€–2π‘₯/(π‘₯^2 + 1) 𝑑π‘₯+γ€— 1/5 ∫1β–’1/(π‘₯^2 + 1) 𝑑π‘₯ 𝐈𝟐 1/5 ∫1β–’2π‘₯/(π‘₯^2+ 1) 𝑑π‘₯ Let 𝑑=π‘₯^2+ 1 𝑑𝑑/𝑑π‘₯=2π‘₯ 𝑑𝑑=2π‘₯ 𝑑π‘₯ Substituting, =1/5 ∫1▒𝑑𝑑/𝑑 = 1/5 log |𝑑| + C_2 = 1/5 log |π‘₯^2+1| + C_2 πˆπŸ‘ 1/5 ∫1β–’1/(π‘₯^2+ 1) 𝑑π‘₯ = 1/5 γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (π‘₯)+C_3 Hence ∫1β–’(π‘₯^2+ π‘₯ + 1)/((π‘₯ + 2) (π‘₯^2+ 1)) 𝑑π‘₯ =πŸ‘/πŸ“ π’π’π’ˆ|𝒙+𝟐|+𝟏/πŸ“ π’π’π’ˆ|𝒙^𝟐+𝟏|+𝟏/πŸ“ 〖𝒕𝒂𝒏〗^(βˆ’πŸ) (𝒙)+ C where C = C_1+ C_2+C_3

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.