Ex 7.4, 23 - Integrate 5x + 3 / root x^2 + 4x + 10 - Teachoo

Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.4, 23 - Chapter 7 Class 12 Integrals - Part 7

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo