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Example 25 Evaluate the following integrals: (iii) ∫_1^2▒(𝑥 𝑑𝑥)/((𝑥 + 1) (𝑥 + 2) ) 𝑑𝑥 𝐹(𝑥)=∫1▒(𝑥 𝑑𝑥)/(𝑥 + 1)(𝑥 + 2) We can write the integrate as : 𝑥/(𝑥 + 1)(𝑥 + 2) =A/(𝑥 + 1)+B/(𝑥 + 2) 𝑥/(𝑥 + 1)(𝑥 + 2) =(A(𝑥 + 2) + B(𝑥 + 1))/(𝑥 + 1)(𝑥 + 2) Canceling denominators 𝑥=A(𝑥+2)+B(𝑥+1) Putting 𝒙=−𝟐 −2=A(2+2)+B(2+1) −2=A ×0+B(−1) −2=−B B=2 Putting 𝒙=−𝟏 −1=A(−1+2)+B(−1+1) −1=A(1)+B(0) −1=A A=−1 =−𝑙𝑜𝑔|𝑥+1|+𝑙𝑜𝑔|𝑥+2|^2 =𝑙𝑜𝑔|𝑥+2|^2−𝑙𝑜𝑔|𝑥+1| =𝑙𝑜𝑔|(𝑥 + 2)^2/(𝑥 + 1)| Hence 𝐹(𝑥)=𝑙𝑜𝑔|(𝑥 + 2)^2/(𝑥 + 1)| Now, ∫_1^2▒〖𝑥/(𝑥 + 1)(𝑥 + 2) 𝑑𝑥〗=𝐹(2)−𝐹(1) ∫_1^2▒〖𝑥/(𝑥 + 1)(𝑥 + 2) 𝑑𝑥〗=𝑙𝑜𝑔|(2 + 2)^2/(2 + 1)|−𝑙𝑜𝑔|(1 + 2)^2/(1 + 1)| =𝑙𝑜𝑔|(4)^2/3|−𝑙𝑜𝑔|(3)^2/2| =𝑙𝑜𝑔|(4^2/3)/(3^2/2)| =𝑙𝑜𝑔|4^2/3 × 2/3^2 | =𝑙𝑜𝑔|16/3 × 2/9| =𝑙𝑜𝑔|32/27 | =𝐥𝐨𝐠⁡(𝟑𝟐/𝟐𝟕)

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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 and Computer Science at Teachoo