Ex 7.4

Chapter 7 Class 12 Integrals
Serial order wise

### 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