# Ex 7.9, 16

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex7.9, 16 12 5 𝑥2 𝑥2 + 4𝑥 + 3 𝑑𝑥 Step 1 :- Let F 𝑥= 5 𝑥2 𝑥2 + 4𝑥 + 3 𝑑𝑥 =5 𝑥2 𝑥2 + 4𝑥 + 3 𝑑𝑥 =5 𝑥2 + 4𝑥 + 3 − 4𝑥 − 3 𝑥2 + 4𝑥 + 3 𝑑𝑥 =5 𝑥2 + 4𝑥 + 3 𝑥2 + 4𝑥 + 3 𝑑𝑥−5 (4𝑥 + 3) 𝑥2 + 4𝑥 + 3𝑑𝑥 = 5− 20𝑥 + 15 𝑥2 + 4𝑥 + 3 𝑑𝑥 We can write the integrate 20𝑥 + 15 𝑥2 + 4𝑥 + 3= 20𝑥 + 15 𝑥 + 3 𝑥 + 1 Let 20𝑥 + 15 𝑥+3 𝑥+1= A𝑥 + 3+ B𝑥 + 1 20𝑥 + 15 𝑥+3 𝑥+1= A 𝑥 + 1 + B 𝑥 + 3 𝑥+3 𝑥+1 By 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 Similarly Putting 𝑥=−3 20 −3+15=A −3+1+B −3+3 −60+15=A −2 B×0 −45=−2A A= 452 Hence 5 𝑥2 𝑥2 + 4𝑥 + 3 = 5− 20𝑥 + 15 𝑥2 + 4𝑥 + 3 𝑑𝑥 = 5− A𝑥+3− B𝑥+1𝑑𝑥 = 5 𝑑𝑥− 452𝑥 +_3𝑑𝑥− −5 2𝑥+1 𝑑𝑥 =5𝑥− 452𝑙𝑜𝑔 𝑥+3+ 52𝑙𝑜𝑔 𝑥+1 Hence F 𝑥=5𝑥− 52 9 𝑙𝑜𝑔 𝑥+3−𝑙𝑜𝑔 𝑥+1 Step 2 :- 12 5 𝑥2 𝑥2 + 4𝑥 + 3𝑑𝑥=𝐹 2−𝐹 1 =5×2− 52 9 𝑙𝑜𝑔 2+3−𝑙𝑜𝑔 2+1 − 5 ×1+ 529 𝑙𝑜𝑔 1+3−𝑙𝑜𝑔 1+1 =10− 52 9 𝑙𝑜𝑔 5−𝑙𝑜𝑔 3−5− 52 9 𝑙𝑜𝑔 4−𝑙𝑜𝑔 2 =10−5− 52 9𝑙𝑜𝑔 5−9𝑙𝑜𝑔 4−(𝑙𝑜𝑔 3−𝑙𝑜𝑔 2) =𝟓− 𝟓𝟐 𝟗 𝐥𝐨𝐠 𝟓𝟒−𝐥𝐨𝐠 𝟑𝟐

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.