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Example 10 - Find integrals (i) x + 2 / 2x2 + 6x + 5 dx - Examples

Example 10 - Chapter 7 Class 12 Integrals - Part 2
Example 10 - Chapter 7 Class 12 Integrals - Part 3 Example 10 - Chapter 7 Class 12 Integrals - Part 4 Example 10 - Chapter 7 Class 12 Integrals - Part 5 Example 10 - Chapter 7 Class 12 Integrals - Part 6 Example 10 - Chapter 7 Class 12 Integrals - Part 7

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Example 10 Find the following integrals: (i) ∫1β–’(π‘₯ + 2)/(2π‘₯^2 + 6π‘₯ + 5 ) 𝑑π‘₯ We can write numerator as π‘₯+2= A 𝑑/𝑑π‘₯ (2π‘₯^2+6π‘₯+5) + B π‘₯+2= A [4π‘₯+6]+ B π‘₯+2=4𝐴π‘₯+6A+B Finding A & B Comparing coefficient of π‘₯ π‘₯=4𝐴π‘₯ 1 =4A A=1/4 Comparing constant term 2=6A+B 2=6(1/4)+B 2=3/2+B B=2βˆ’3/2=1/2 Now, we know that π‘₯+2= A [4π‘₯+6]+ B π‘₯+2=1/4 [4π‘₯+6]+1/2 Now, our equation is ∫1β–’γ€–(π‘₯ + 2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯=∫1β–’γ€–(1/4 [4π‘₯ + 6] + 1/2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€—γ€— =∫1β–’γ€–(1/4 [4π‘₯ + 6])/(2π‘₯^2 + 6π‘₯ + 5)+∫1β–’γ€–(1/2)/(2π‘₯^2+6π‘₯+5).𝑑π‘₯γ€—γ€— =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5) 𝑑π‘₯+1/2 ∫1β–’γ€–1/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€—γ€— Solving I1 I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5) 𝑑π‘₯γ€— Let t = 2π‘₯^2 + 6π‘₯ + 5 Differentiating both sides w.r.t.π‘₯ 4π‘₯ +6=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(4π‘₯ + 6) Now, I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€— Putting the values of (2π‘₯^2+6π‘₯+5) and 𝑑π‘₯, we get I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/𝑑.𝑑𝑑/(4π‘₯ + 6) γ€— I1 =1/4 ∫1β–’γ€–1/𝑑.𝑑𝑑 γ€— I1 =1/4 π‘™π‘œπ‘”|𝑑|+𝐢1 I1 =1/4 π‘™π‘œπ‘”|2π‘₯^2+6π‘₯+5|+𝐢1 Solving I2 I2 =1/2 ∫1β–’γ€–1/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯ γ€— I2 =1/2 ∫1β–’γ€–1/2[π‘₯^2 + 6π‘₯/2 + 5/2 ] .𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 +3π‘₯ + 5/2).𝑑π‘₯ γ€— (Using ∫1β–’γ€–1/π‘₯.𝑑π‘₯=π‘™π‘œπ‘”|π‘₯|+𝐢1γ€—) (Using 𝑑=2π‘₯^2+6π‘₯+5) I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 + 2(π‘₯)(3/2) + 5/2).𝑑π‘₯ γ€— Adding & subtracting (3/2)^2 in denominator I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 + 2(π‘₯) γ€–(3/2) +(3/2)^2βˆ’ (3/2)γ€—^2 + 5/2).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2 βˆ’ (3/2)^2 + 5/2) 𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2 βˆ’ 9/4 + 5/2).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ (βˆ’9 + 10)/4 ).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ 1/4 ).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ (1/2)^2 ).𝑑π‘₯ γ€— =1/4 [1/(1/2) tan^(βˆ’1)⁑〖(π‘₯ + 3/2)/(1/2)+𝐢2γ€— ] =1/4 [2 tan^(βˆ’1)⁑〖((2π‘₯ + 3)/2)/(1/2)+𝐢2γ€— ] =1/4 [2 γ€–tan^(βˆ’1) (2π‘₯+3)〗⁑〖+𝐢2γ€— ] It is of form ∫1▒〖𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯/π‘Ž+𝐢2" " γ€— γ€— Replacing π‘₯ by (π‘₯+3/2) and by 1/2, we get =2/4 tan^(βˆ’1)⁑〖(2π‘₯+3)+𝐢2/4γ€— =1/2 tan^(βˆ’1)⁑〖(2π‘₯+3)+𝐢3γ€— Now, putting the value of I1 and I2 in eq. (1) ∴ ∫1β–’γ€–(π‘₯+2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€— =1/4 π‘™π‘œπ‘”|2π‘₯^2+6π‘₯+5|+𝐢1+1/2 tan^(βˆ’1)⁑〖(2π‘₯+3)+γ€— 𝐢3 =𝟏/πŸ’ π’π’π’ˆ|πŸπ’™^𝟐+πŸ”π’™+πŸ“|+𝟏/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖(πŸπ’™+πŸ‘)+γ€— π‘ͺ (where 𝐢3 = 𝐢2/4 )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.