Ex 7.9, 16 - Chapter 7 Class 12 Integrals (Term 2)

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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)