Ex 7.4, 23 - Chapter 7 Class 12 Integrals (Term 2)

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Ex 7.4, 23 Integrate (5𝑥 + 3)/√(𝑥^2 + 4𝑥 + 10) ∫1▒(5𝑥 + 3)/√(𝑥^2 + 4𝑥 + 10) 𝑑𝑥 =5∫1▒(𝑥 + 3/5)/√(𝑥^2 + 4𝑥 + 10) =5/2 ∫1▒(2𝑥 + 6/5)/√(𝑥^2 + 4𝑥 + 10) =5/2 ∫1▒(2𝑥 + 4 − 4 + 6/5)/√(𝑥^2 + 4𝑥 + 10) =5/2 ∫1▒(2𝑥 + 4 − 14/5)/√(𝑥^2 + 4𝑥 + 10) Rough (𝑥^2+4𝑥+10)^′=2𝑥+4 =5/2 ∫1▒(2𝑥 + 4)/√(𝑥^2 + 4𝑥 + 10) 𝑑𝑥+5/2 ∫1▒(− 14/5)/√(𝑥^2 + 4𝑥 + 10) 𝑑𝑥 =5/2 ∫1▒(2𝑥 + 4)/√(𝑥^2 + 4𝑥 + 10) 𝑑𝑥−7∫1▒𝑑𝑥/√(𝑥^2 + 4𝑥 + 10) 𝑑𝑥 Solving 𝑰𝟏 I1=5/2 ∫1▒( 2𝑥 + 4)/√(𝑥^2 + 4𝑥 + 10) . 𝑑𝑥 Let 𝑥^2 + 4𝑥 + 10=𝑡 Diff both sides w.r.t.x 2𝑥+4+0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(2𝑥 + 4) Thus, our equation becomes I1=5/2 ∫1▒( 2𝑥 + 4)/√(𝑥^2 + 4𝑥 + 10) . 𝑑𝑥 Putting the value of (𝑥^2+4𝑥+10)=𝑡 and 𝑑𝑥=𝑑𝑡/(2𝑥 + 4) I1=5/2 ∫1▒(2𝑥 + 4)/√𝑡 . 𝑑𝑥 I1=5/2 ∫1▒(2𝑥 + 4)/√𝑡 .𝑑𝑡/(2𝑥 + 4) I1=5/2 ∫1▒1/√𝑡 . 𝑑𝑡 I1=5/2 ∫1▒1/(𝑡)^(1/2) . 𝑑𝑡 I1=5/2 ∫1▒(𝑡)^((− 1)/2) . 𝑑𝑡 I1=5/2 〖𝑡 〗^((−1)/2 + 1)/((−1)/2 + 1) +𝐶1 I1=5/2 (𝑡 ^(1/2 ))/(1/2) +𝐶1 I1=5 𝑡 ^(1/2 )+𝐶1 I1=5 √𝑡+𝐶1 I1=5 √(𝑥^2+4𝑥+10)+𝐶1 (Using 𝑡=𝑥^2+4𝑥+1) Solving 𝑰𝟐 I2=∫1▒( 7)/√(𝑥^2 + 4𝑥 + 10) . 𝑑𝑥 I2=7∫1▒1/√(𝑥^2 + 2(2)(𝑥) + 10) . 𝑑𝑥 I2=7∫1▒1/√(𝑥^2 + 2(2)(𝑥) + (2)^2 − (2)^2 + 10) . 𝑑𝑥 I2=7∫1▒1/√((𝑥 + 2)^2 − (2)^2 + 10) . 𝑑𝑥 I2=7∫1▒1/√((𝑥 + 2)^2 − 4 + 10) . 𝑑𝑥 I2=7∫1▒1/√((𝑥 + 2)^2 + 6) . 𝑑𝑥 I2=7∫1▒1/√((𝑥 + 2)^2 + (√6 )^2 ) . 𝑑𝑥 I2=7[𝑙𝑜𝑔⁡|𝑥+2+√((𝑥+2)^2 + (√6)^2 )| ]+𝐶2 I2=7 𝑙𝑜𝑔⁡|𝑥+2+√(𝑥^2+4𝑥+4+6)|+𝐶2 I2=7 𝑙𝑜𝑔⁡|𝑥+2+√(𝑥^2+4𝑥+10)|+𝐶2 It is of form ∫1▒𝑑𝑥/√(𝑥^2 + 𝑎^2 ) =𝑙𝑜𝑔⁡|𝑥+√(𝑥^2 + 𝑎^2 )|+𝐶2 ∴ Replacing x by (𝑥+2) and a by √6 , we get Putting the values of I1 and I2 in (1) ∫1▒(5𝑥 + 3)/√(𝑥^2 + 4𝑥 + 10) . 𝑑𝑥 = 𝐼_1−𝐼_2 =5 √(𝑥^2+4𝑥+10)+𝐶1−7 𝑙𝑜𝑔⁡|𝑥+2+√(𝑥^2+4𝑥+10)|+𝐶2 =𝟓 √(𝒙^𝟐+𝟒𝒙+𝟏𝟎)−𝟕 𝒍𝒐𝒈⁡|𝒙+𝟐+√(𝒙^𝟐+𝟒𝒙+𝟏𝟎)|+𝑪