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Ex 7.4, 21 - Integrate x + 2 / root x^2 + 2x + 3 - Chapter 7 Class 12

Ex 7.4, 21 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 21 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.4, 21 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.4, 21 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.4, 21 - Chapter 7 Class 12 Integrals - Part 6


Transcript

Ex 7.4, 21 Integrate the function (𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) ∫1▒(𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥 =1/2 ∫1▒(2𝑥 + 4)/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥 =1/2 ∫1▒(2𝑥 + 2 + 4 − 2)/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥 =1/2 ∫1▒(2𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥+2/2 ∫1▒𝑑𝑥/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥 =1/2 ∫1▒(2𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥+∫1▒𝑑𝑥/√(𝑥^2 + 2𝑥 + 3) 𝑑𝑥 Rough (𝑥^2+2𝑥+3)^′=2𝑥+2 Solving 𝑰𝟏 I1=1/2 ∫1▒(2𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) . 𝑑𝑥 Let 𝑥^2 + 2𝑥 + 3=𝑡 Diff both sides w.r.t.x 2𝑥+2+0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(2𝑥 + 2) Now, our equation becomes I1=1/2 ∫1▒(2𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3) . 𝑑𝑥 Putting the value of (4𝑥−𝑥^2 ) and 𝑑𝑥 I1=1/2 ∫1▒(2𝑥 + 2)/√𝑡 . 𝑑𝑥 I1=1/2 ∫1▒(2𝑥 + 2)/√𝑡 . 𝑑𝑡/(2𝑥 + 2) I1=1/2 ∫1▒1/√𝑡 . 𝑑𝑡 I1=1/2 ∫1▒1/(𝑡)^(1/2) . 𝑑𝑡 I1=1/2 ∫1▒(𝑡)^((− 1)/2) . 𝑑𝑡 I1=1/2 〖𝑡 〗^((−1)/2 + 1)/((−1)/2 + 1) +𝐶1 I1= 𝑡 ^(1/2 )+𝐶1 I1= √𝑡+𝐶1 I1=√(𝑥^2+2𝑥+3)+𝐶 Solving 𝑰𝟐 I2=∫1▒1/√(𝑥^2 + 2𝑥 + 3) . 𝑑𝑥 I2=∫1▒1/√(𝑥^2 + 2(𝑥)(1) + 3) . 𝑑𝑥 I2=∫1▒1/√(𝑥^2 + 2(𝑥)(1) − (1)^2 + (1)^2 + 3) . 𝑑𝑥 (Using 𝑡=𝑥^2+2𝑥+3) I2=∫1▒1/√((𝑥 + 1)^2 − (1)^2 + 3) . 𝑑𝑥 I2=∫1▒1/√((𝑥 + 1)^2 − 1 + 3) . 𝑑𝑥 I2=∫1▒1/√((𝑥 + 1)^2 + 2) . 𝑑𝑥 I2=∫1▒1/√((𝑥 + 1)^2 +(√2 )^2 ) . 𝑑𝑥 I2=𝑙𝑜𝑔⁡|𝑥+1+√((𝑥 + 1)^2+(√2 )^2 )|+𝐶2 It is of form ∫1▒𝑑𝑥/√(𝑥^2 + 𝑎^2 ) =𝑙𝑜𝑔⁡|𝑥+√(𝑥^2 + 𝑎^2 )|+𝐶2 ∴ Replacing x by (𝑥 + 1) and a by √2 , we get I2=𝑙𝑜𝑔⁡|𝑥+1+√(𝑥^2+2𝑥+1+2)|+𝐶2 I2=𝑙𝑜𝑔⁡|𝑥+1+√(𝑥^2+2𝑥+3)|+𝐶2 Putting the values of I1 and I2 in (1) ∫1▒〖(𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3).〗 . 𝑑𝑥 = 𝐼_1+𝐼_2 =√(𝑥^2+2𝑥+3)+𝐶1+𝑙𝑜𝑔⁡|𝑥+1+√(𝑥^2+2𝑥+3)|+𝐶2 =√(𝒙^𝟐+𝟐𝒙+𝟑)+𝒍𝒐𝒈⁡|𝒙+𝟏+√(𝒙^𝟐+𝟐𝒙+𝟑)|+𝑪

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.