Ex 7.8

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.8, 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) −5=2B 𝐁=(−𝟓)/( 𝟐) 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)

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