Evaluate Integral ∫ x dx / (x + 1) (x + 2) from 1

integral of 1/(1+sqrt(x))^4, calculus u-substitution
integral of 1/(1+sqrt(x))^4, calculus u-substitution

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Question 1 Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at June 13, 2023 by Teachoo

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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