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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.