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Example 10 Find the following integrals: (ii) ∫1β–’(π‘₯ + 3)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ We can write numerator as π‘₯+3= A𝑑/𝑑π‘₯ [βˆ’π‘₯^2βˆ’4π‘₯+5]+ B π‘₯+3= A[βˆ’2π‘₯βˆ’4]+ B π‘₯+3=βˆ’2𝐴π‘₯βˆ’4A+B Finding A & B Comparing coefficient of π‘₯ π‘₯=βˆ’2𝐴π‘₯ 1 =βˆ’2A A=(βˆ’1)/( 2) Comparing constant term 3=βˆ’4A+B 3=βˆ’4((βˆ’1)/( 2))+B 3=2+B B=3βˆ’2=1 Now, we know that π‘₯+3=A[βˆ’2π‘₯βˆ’4]+B π‘₯+3=(βˆ’1)/( 2) [βˆ’2π‘₯βˆ’4]+1 Now, our equation is ∫1β–’(π‘₯+3)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ).𝑑π‘₯=∫1β–’(1/2 [βˆ’2π‘₯ βˆ’ 4]+1)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯ =∫1β–’((βˆ’1)/2 [βˆ’2π‘₯ βˆ’ 4]+1)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ) 𝑑π‘₯+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ =(βˆ’1)/2 ∫1β–’(βˆ’2π‘₯ βˆ’ 4)/√(5 βˆ’ 4π‘₯ + π‘₯^2 ) 𝑑π‘₯+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ Solving I1 I1 = (βˆ’1)/2 ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/√(βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5).𝑑π‘₯γ€— Let t = βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5 Differentiating both sides w.r.t. π‘₯ βˆ’2π‘₯ βˆ’ 4 =𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(βˆ’ 2π‘₯ βˆ’ 4) Now, I1 = (βˆ’1)/( 2) ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/√(βˆ’π‘₯^2 βˆ’ 4π‘₯ + 5).𝑑π‘₯γ€— Putting the values of (βˆ’π‘₯^2βˆ’4π‘₯+5) and 𝑑π‘₯ I1 = (βˆ’1)/( 2) ∫1β–’γ€–(βˆ’2π‘₯ βˆ’ 4)/βˆšπ‘‘.𝑑𝑑/((βˆ’2π‘₯ βˆ’4) )γ€— I1 = (βˆ’1)/( 2) ∫1β–’γ€–1/βˆšπ‘‘.𝑑𝑑〗 I1 = (βˆ’1)/( 2) ∫1β–’γ€–1/(𝑑)^(1/2) .𝑑𝑑〗 I1 = (βˆ’1)/( 2) ∫1β–’γ€–(𝑑)^((βˆ’1)/2) 𝑑𝑑〗 I1 = (βˆ’1)/( 2) 𝑑^((βˆ’1)/( 2) + 1)/(((βˆ’1)/( 2) + 1) )+𝐢1 I1 = (βˆ’1)/( 2) 𝑑^(1/( 2) )/((1/2) )+𝐢1 I1 = βˆ’ 𝑑^(1/( 2) )+𝐢1 I1 = βˆ’βˆšπ‘‘+𝐢1 Putting back 𝑑=π‘₯^2βˆ’4π‘₯+5) I1 = βˆ’βˆš(βˆ’π‘₯^2βˆ’4π‘₯+5)+𝐢1 Solving π‘°πŸ I2 = ∫1β–’γ€–1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’(π‘₯^2 + 4π‘₯ βˆ’ 5) ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[π‘₯^2 + 2(2)(π‘₯) βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[π‘₯^2 + 2(2)(π‘₯) + (2)^2 βˆ’ (2)^2 βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[(π‘₯ + 2)^2 βˆ’ 4 βˆ’ 5] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(βˆ’[(π‘₯ + 2)^2 βˆ’ 9] ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√(9 βˆ’ (π‘₯ + 2)^2 ) 𝑑π‘₯γ€— I2 = ∫1β–’γ€–1/√((3)^2 βˆ’ (π‘₯ + 2)^2 ) 𝑑π‘₯γ€— It is of form ∫1β–’γ€–1/√(π‘Ž^2 βˆ’ π‘₯^2 ).𝑑π‘₯=〖𝑠𝑖𝑛〗^(βˆ’1)⁑〖π‘₯/π‘Ž+𝐢2γ€— γ€— Replacing π‘₯ by (π‘₯+2) and a by 3, we get I2 = sin^(βˆ’1)⁑〖((π‘₯ + 2)/3)+𝐢2γ€— Now, Putting values of I1 and I2 in (1) ∫1β–’γ€–(π‘₯ + 3)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ).𝑑π‘₯=(βˆ’1)/( 2) ∫1β–’(βˆ’ 2π‘₯ βˆ’ 4)/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 )γ€—+∫1β–’1/√(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 ) 𝑑π‘₯ =βˆ’βˆš(5 βˆ’ 4π‘₯ βˆ’ π‘₯^2 )+𝐢1+sin^(βˆ’1)⁑((π‘₯ + 2)/3)+𝐢2 =βˆ’βˆš(πŸ“ βˆ’ πŸ’π’™ βˆ’ 𝒙^𝟐 )+γ€–π’”π’Šπ’γ€—^(βˆ’πŸ)⁑〖((𝒙 + 𝟐)/πŸ‘)+π‘ͺγ€—

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.