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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 13 Find ∫1β–’(3π‘₯ βˆ’2)/((π‘₯ + 1)^2 (π‘₯ + 3) ) 𝑑π‘₯ We can write Integral as (3π‘₯ βˆ’ 2)/((π‘₯ + 1)^2 (π‘₯ + 3) )=𝐴/(π‘₯ + 1) + 𝐡/(π‘₯ + 1)^2 + 𝐢/((π‘₯ + 3) ) (3π‘₯ βˆ’ 2)/((π‘₯ + 1)^2 (π‘₯ + 3) )=(𝐴(π‘₯ + 1)(π‘₯ + 3) + 𝐡(π‘₯ + 3) + 𝐢(π‘₯ + 1)^2)/((π‘₯ + 1)^2 (π‘₯ + 3) ) Cancelling denominator 3π‘₯ βˆ’2=𝐴(π‘₯+1)(π‘₯+3)+𝐡(π‘₯+3)+𝐢(π‘₯+1)^2 Putting x = βˆ’1 3(βˆ’1) βˆ’2=𝐴(βˆ’1+1)(βˆ’1+3)+𝐡(βˆ’1+3)+𝐢(βˆ’1+1)^2 βˆ’3βˆ’2=𝐴×0+𝐡×2+𝐢×(0)^2 βˆ’5=𝐡×2 𝐡=(βˆ’ 5)/2 Putting x = βˆ’ 3 3π‘₯ βˆ’2=𝐴(π‘₯+1)(π‘₯+3)+𝐡(π‘₯+3)+𝐢(π‘₯+1)^2 3(βˆ’3)βˆ’2=𝐴(βˆ’3+1)(βˆ’3+3)+𝐡(βˆ’3+3)+𝐢(βˆ’3+1)^2 βˆ’9βˆ’2=𝐴×0+𝐡×0+𝐢×(βˆ’2)^2 βˆ’11=0+0+𝐢(4) βˆ’11=4𝐢 (βˆ’11)/4 =𝐢 𝐢 =(βˆ’11)/4 Putting x = 0 3π‘₯ βˆ’2=𝐴(π‘₯+1)(π‘₯+3)+𝐡(π‘₯+3)+𝐢(π‘₯+1)^2 3(0) βˆ’ 2 = A(1) (3) + B(3) + C γ€–"(1)" γ€—^2 βˆ’2 = 3A + 3B + C Putting value of B & C βˆ’2 = 3A + 3((βˆ’5)/2) + ((βˆ’11)/4) βˆ’2 = 3A βˆ’ 15/2βˆ’11/4 βˆ’2 = 3A + (βˆ’30 βˆ’ 11)/4 βˆ’8 = 12A βˆ’ 41 41 βˆ’ 8 = 12A 33/12 = A 11/4 = A Hence, we can write (3π‘₯ βˆ’ 2)/((π‘₯ + 1)^2 (π‘₯ + 3) )=𝐴/(π‘₯ + 1) + 𝐡/(π‘₯ + 1)^2 + 𝐢/((π‘₯ + 3) ) (3π‘₯ βˆ’ 2)/((π‘₯ + 1)^2 (π‘₯ + 3) )=11/4(π‘₯ + 1) βˆ’ 5/γ€–2(π‘₯ + 1)γ€—^2 βˆ’ 11/4(π‘₯ + 3) Therefore ∫1β–’(3π‘₯ βˆ’ 2)/((π‘₯ + 1)^2 (π‘₯ + 3) ) 𝑑π‘₯ =∫1β–’11/4(π‘₯ + 1) 𝑑π‘₯βˆ’βˆ«1β–’5/γ€–2(π‘₯ + 1)γ€—^2 𝑑π‘₯βˆ’βˆ«1β–’11/4(π‘₯ + 3) 𝑑π‘₯ =11/4 log⁑|π‘₯+1|βˆ’5/2Γ—((βˆ’1))/((π‘₯ + 1) ) βˆ’ 11/4 log⁑|π‘₯+3|+𝐢 =11/4 (log⁑|π‘₯+1|βˆ’log⁑|π‘₯+3| )+5/(2 (π‘₯ + 1) )+𝐢 =𝟏𝟏/πŸ’ π’π’π’ˆβ‘|(𝒙 + 𝟏)/(𝒙 + πŸ‘)| + πŸ“/(𝟐 (𝒙 + 𝟏) )+π‘ͺ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.