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) ∫1▒(𝑥 + 2)/(2𝑥^2 + 6𝑥 + 5 ) 𝑑𝑥 We can write numerator as 𝑥+2= A 𝑑/𝑑𝑥 (2𝑥^2+6𝑥+5) + B 𝑥+2= A [4𝑥+6]+ B 𝑥+2=4𝐴𝑥+6A+B Finding A & B Comparing coefficient of 𝑥 𝑥=4𝐴𝑥 1 =4A A=1/4 Comparing constant term 2=6A+B 2=6(1/4)+B 2=3/2+B B=2−3/2=1/2 Now, we know that 𝑥+2= A [4𝑥+6]+ B 𝑥+2=1/4 [4𝑥+6]+1/2 Now, our equation is ∫1▒〖(𝑥 + 2)/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥=∫1▒〖(1/4 [4𝑥 + 6] + 1/2)/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥〗〗 =∫1▒〖(1/4 [4𝑥 + 6])/(2𝑥^2 + 6𝑥 + 5)+∫1▒〖(1/2)/(2𝑥^2+6𝑥+5).𝑑𝑥〗〗 =1/4 ∫1▒〖(4𝑥 + 6)/(2𝑥^2 + 6𝑥 + 5) 𝑑𝑥+1/2 ∫1▒〖1/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥〗〗 Solving I1 I1 =1/4 ∫1▒〖(4𝑥 + 6)/(2𝑥^2 + 6𝑥 + 5) 𝑑𝑥〗 Let t = 2𝑥^2 + 6𝑥 + 5 Differentiating both sides w.r.t.𝑥 4𝑥 +6=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(4𝑥 + 6) Now, I1 =1/4 ∫1▒〖(4𝑥 + 6)/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥〗 Putting the values of (2𝑥^2+6𝑥+5) and 𝑑𝑥, we get I1 =1/4 ∫1▒〖(4𝑥 + 6)/𝑡.𝑑𝑡/(4𝑥 + 6) 〗 I1 =1/4 ∫1▒〖1/𝑡.𝑑𝑡 〗 I1 =1/4 𝑙𝑜𝑔|𝑡|+𝐶1 I1 =1/4 𝑙𝑜𝑔|2𝑥^2+6𝑥+5|+𝐶1 Solving I2 I2 =1/2 ∫1▒〖1/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥 〗 I2 =1/2 ∫1▒〖1/2[𝑥^2 + 6𝑥/2 + 5/2 ] .𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/(𝑥^2 +3𝑥 + 5/2).𝑑𝑥 〗 (Using ∫1▒〖1/𝑥.𝑑𝑥=𝑙𝑜𝑔|𝑥|+𝐶1〗) (Using 𝑡=2𝑥^2+6𝑥+5) I2 =1/4 ∫1▒〖1/(𝑥^2 + 2(𝑥)(3/2) + 5/2).𝑑𝑥 〗 Adding & subtracting (3/2)^2 in denominator I2 =1/4 ∫1▒〖1/(𝑥^2 + 2(𝑥) 〖(3/2) +(3/2)^2− (3/2)〗^2 + 5/2).𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/((𝑥 + 3/2)^2 − (3/2)^2 + 5/2) 𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/((𝑥 + 3/2)^2 − 9/4 + 5/2).𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/((𝑥 + 3/2)^2+ (−9 + 10)/4 ).𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/((𝑥 + 3/2)^2+ 1/4 ).𝑑𝑥 〗 I2 =1/4 ∫1▒〖1/((𝑥 + 3/2)^2+ (1/2)^2 ).𝑑𝑥 〗 =1/4 [1/(1/2) tan^(−1)⁡〖(𝑥 + 3/2)/(1/2)+𝐶2〗 ] =1/4 [2 tan^(−1)⁡〖((2𝑥 + 3)/2)/(1/2)+𝐶2〗 ] =1/4 [2 〖tan^(−1) (2𝑥+3)〗⁡〖+𝐶2〗 ] It is of form ∫1▒〖𝑑𝑥/(𝑥^2 + 𝑎^2 )=1/𝑎 〖𝑡𝑎𝑛〗^(−1)⁡〖𝑥/𝑎+𝐶2" " 〗 〗 Replacing 𝑥 by (𝑥+3/2) and by 1/2, we get =2/4 tan^(−1)⁡〖(2𝑥+3)+𝐶2/4〗 =1/2 tan^(−1)⁡〖(2𝑥+3)+𝐶3〗 Now, putting the value of I1 and I2 in eq. (1) ∴ ∫1▒〖(𝑥+2)/(2𝑥^2 + 6𝑥 + 5).𝑑𝑥〗 =1/4 𝑙𝑜𝑔|2𝑥^2+6𝑥+5|+𝐶1+1/2 tan^(−1)⁡〖(2𝑥+3)+〗 𝐶3 =𝟏/𝟒 𝒍𝒐𝒈|𝟐𝒙^𝟐+𝟔𝒙+𝟓|+𝟏/𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖(𝟐𝒙+𝟑)+〗 𝑪 (where 𝐶3 = 𝐶2/4 )

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.