Ex 7.9, 16

Transcript

Ex7.9, 16 1﷮2﷮ 5 𝑥﷮2﷯﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯﷯ 𝑑𝑥 Step 1 :- Let F 𝑥﷯= ﷮﷮ 5 𝑥﷮2﷯﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯ 𝑑𝑥﷯ =5 ﷮﷮ 𝑥﷮2﷯﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯ 𝑑𝑥﷯ =5 ﷮﷮ 𝑥﷮2﷯ + 4𝑥 + 3 − 4𝑥 − 3﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯ 𝑑𝑥﷯ =5 ﷮﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯ 𝑑𝑥﷯−5 ﷮﷮ (4𝑥 + 3)﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯𝑑𝑥﷯ = ﷮﷮ 5− 20𝑥 + 15 ﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯﷯ 𝑑𝑥﷯ We can write the integrate 20𝑥 + 15﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯﷯= 20𝑥 + 15﷮ 𝑥 + 3﷯ 𝑥 + 1﷯﷯ 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﷯﷯ 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= 45﷮2﷯ Hence ﷮﷮ 5 𝑥﷮2﷯﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯ = ﷮﷮ 5− 20𝑥 + 15 ﷮ 𝑥﷮2﷯ + 4𝑥 + 3﷯﷯ 𝑑𝑥﷯﷯﷯ = ﷮﷮5− A﷮𝑥+3﷯−﷯ B﷮𝑥+1﷯𝑑𝑥 = ﷮﷮5 𝑑𝑥﷯− ﷮﷮ 45﷮2﷯﷮𝑥 +_3﷯𝑑𝑥− ﷮﷮ −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﷯﷯ Step 2 :- 1﷮2﷮ 5 𝑥﷮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 5−9𝑙𝑜𝑔 4−(𝑙𝑜𝑔 3−𝑙𝑜𝑔 2)﷯ =𝟓− 𝟓﷮𝟐﷯ 𝟗 𝐥𝐨𝐠 𝟓﷮𝟒﷯−𝐥𝐨𝐠 𝟑﷮𝟐﷯﷯