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Ex 7.9, 16 - Direct Integrate 5x2 / x2 + 4x + 3 dx from 1 to 2

Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.9, 16 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.9, 16 ∫_1^2▒(5𝑥^2)/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥 Let F(𝑥)=∫1▒〖(5𝑥^2)/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥〗 =5∫1▒〖𝑥^2/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥〗 =5∫1▒〖(𝑥^2 + 4𝑥 + 3 − 4𝑥 − 3)/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥〗 =5∫1▒〖(𝑥^2 + 4𝑥 + 3)/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥〗−5∫1▒〖( (4𝑥 + 3))/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥〗 =∫1▒〖(5−(20𝑥 + 15 )/(𝑥^2 + 4𝑥 + 3)) 𝑑𝑥〗 =∫1▒〖(5−(20𝑥 + 15 )/(𝑥^2 + 3𝑥 + 𝑥 + 3)) 𝑑𝑥〗 =∫1▒〖(5−(20𝑥 + 15 )/(𝑥(𝑥 + 3) + 1(𝑥 + 3))) 𝑑𝑥〗 =∫1▒〖(5−(20𝑥 + 15 )/((𝑥 + 3) (𝑥 + 1))) 𝑑𝑥〗 Now, Let (20𝑥 + 15)/(𝑥 + 3)(𝑥 + 1) =A/(𝑥 + 3)+B/(𝑥 + 1) (20𝑥 + 15)/(𝑥 + 3)(𝑥 + 1) =(A(𝑥 + 1) + B(𝑥 + 3))/(𝑥 + 3)(𝑥 + 1) Canceling denominator 20𝑥+15=A(𝑥 + 1) + B(𝑥 + 3) Putting 𝑥=−1 20(−1)+15=A(−1 + 1) + B(−1 + 3) −20+15=A×0+B (2) Putting 𝑥=−1 20(−1)+15=A(−1 + 1) + B(−1 + 3) −20+15=A×0+B (2) −5=2B B=(−5)/( 2) Similarly Putting 𝑥=−3 20(−3)+15=A(−3+1)+B(−3+3) −60+15=A(−2) B×0 −45=−2A A=45/2 Hence ∫1▒█((5𝑥^2)/(𝑥^2 + 4𝑥 + 3) " " =∫1▒〖(5−(20𝑥 + 15 )/(𝑥^2 + 4𝑥 + 3)) 𝑑𝑥〗) =∫1▒〖5−A/(𝑥+3)−〗 B/(𝑥+1) 𝑑𝑥 =∫1▒〖5 𝑑𝑥〗−∫1▒〖(45/2)/(𝑥 +_3) 𝑑𝑥−∫1▒〖(((−5)/( 2)))/(𝑥+1) 𝑑𝑥〗〗 =5𝑥−45/2 𝑙𝑜𝑔|𝑥+3|+5/2 𝑙𝑜𝑔|𝑥+1| Hence F(𝑥)=5𝑥−5/2 [9 𝑙𝑜𝑔|𝑥+3|−𝑙𝑜𝑔|𝑥+1|] Now, ∫_1^2▒〖(5𝑥^2)/(𝑥^2 + 4𝑥 + 3) 𝑑𝑥=𝐹(2)−𝐹(1) 〗 =[5×2−5/2 (9 𝑙𝑜𝑔|2+3|−𝑙𝑜𝑔|2+1|)] − [5 ×1−5/2 (9 𝑙𝑜𝑔|1+3|−𝑙𝑜𝑔|1+1|)] =10−5/2 [9 𝑙𝑜𝑔 5−𝑙𝑜𝑔 3]−5+5/2 [9 𝑙𝑜𝑔 4−𝑙𝑜𝑔 2] =10−5−5/2 [9𝑙𝑜𝑔 5−𝑙𝑜𝑔 3−9𝑙𝑜𝑔 4+𝑙𝑜𝑔 2)] =10−5−5/2 [9𝑙𝑜𝑔 5−9𝑙𝑜𝑔 4−(𝑙𝑜𝑔 3−𝑙𝑜𝑔 2)] =𝟓−𝟓/𝟐 (𝟗 𝐥𝐨𝐠 𝟓/𝟒−𝐥𝐨𝐠 𝟑/𝟐) (log a − log b = log 𝑎/𝑏)

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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.