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Example 27 - Chapter 7 Class 12 Integrals - Part 7

Example 27 - Chapter 7 Class 12 Integrals - Part 8
Example 27 - Chapter 7 Class 12 Integrals - Part 9
Example 27 - Chapter 7 Class 12 Integrals - Part 10

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Example 27 Evaluate the following integrals: (iii) ∫_1^2β–’(π‘₯ 𝑑π‘₯)/((π‘₯ + 1) (π‘₯ + 2) ) 𝑑π‘₯ 𝐹(π‘₯)=∫1β–’(π‘₯ 𝑑π‘₯)/(π‘₯ + 1)(π‘₯ + 2) We can write the integrate as : π‘₯/(π‘₯ + 1)(π‘₯ + 2) =A/(π‘₯ + 1)+B/(π‘₯ + 2) π‘₯/(π‘₯ + 1)(π‘₯ + 2) =(A(π‘₯ + 2) + B(π‘₯ + 1))/(π‘₯ + 1)(π‘₯ + 2) Canceling denominators π‘₯=A(π‘₯+2)+B(π‘₯+1) Putting 𝒙=βˆ’πŸ βˆ’2=A(2+2)+B(2+1) βˆ’2=A Γ—0+B(βˆ’1) βˆ’2=βˆ’B B=2 Putting 𝒙=βˆ’πŸ βˆ’1=A(βˆ’1+2)+B(βˆ’1+1) βˆ’1=A(1)+B(0) βˆ’1=A A=βˆ’1 Therefore π‘₯/(π‘₯ + 1)(π‘₯ + 2) =(βˆ’1)/(π‘₯ + 1)+2/(π‘₯ + 2) Integrating w.r.t.π‘₯ ∫1β–’γ€–π‘₯/(π‘₯+1)(π‘₯+2) =∫1β–’γ€–(βˆ’1)/((π‘₯+1) ) 𝑑π‘₯γ€—+∫1β–’γ€–2/(π‘₯+2) 𝑑π‘₯γ€—γ€— =βˆ’π‘™π‘œπ‘”|π‘₯+1|+2π‘™π‘œπ‘”|π‘₯+2| =βˆ’π‘™π‘œπ‘”|π‘₯+1|+π‘™π‘œπ‘”|π‘₯+2|^2 =π‘™π‘œπ‘”|π‘₯+2|^2βˆ’π‘™π‘œπ‘”|π‘₯+1| =π‘™π‘œπ‘”|(π‘₯ + 2)^2/(π‘₯ + 1)| Hence 𝐹(π‘₯)=π‘™π‘œπ‘”|(π‘₯ + 2)^2/(π‘₯ + 1)| Now, ∫_1^2β–’γ€–π‘₯/(π‘₯ + 1)(π‘₯ + 2) 𝑑π‘₯γ€—=𝐹(2)βˆ’πΉ(1) ∫_1^2β–’γ€–π‘₯/(π‘₯ + 1)(π‘₯ + 2) 𝑑π‘₯γ€—=π‘™π‘œπ‘”|(2 + 2)^2/(2 + 1)|βˆ’π‘™π‘œπ‘”|(1 + 2)^2/(1 + 1)| (𝑏 logβ‘γ€–π‘Ž=logβ‘γ€–π‘Ž^𝑏 γ€— γ€— ) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡) =π‘™π‘œπ‘”|(4)^2/3|βˆ’π‘™π‘œπ‘”|(3)^2/2| =π‘™π‘œπ‘”|(4^2/3)/(3^2/2)| =π‘™π‘œπ‘”|4^2/3 Γ— 2/3^2 | =π‘™π‘œπ‘”|16/3 Γ— 2/9| =π‘™π‘œπ‘”|32/27 | =π₯𝐨𝐠⁑(πŸ‘πŸ/πŸπŸ•) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡) (log⁑Aβˆ’log⁑B=π‘™π‘œπ‘” A/𝐡)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.