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Ex 7.4, 22 - Integrate x + 3 / x^2 - 2x - 5 - Class 12 NCERT

Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 4 Ex 7.4, 22 - Chapter 7 Class 12 Integrals - Part 5

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Ex 7.4, 22 Integrate the function (𝑥 + 3)/(𝑥^2 − 2𝑥 − 5) ∫1▒(𝑥 + 3)/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥 =1/2 ∫1▒(2𝑥 + 6)/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥 =1/2 ∫1▒(2𝑥 − 2 + 2 + 6 )/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥 =1/2 ∫1▒(2𝑥 − 2)/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥+8/2 ∫1▒𝑑𝑥/(𝑥^2 − 2𝑥 − 5) =1/2 ∫1▒(2𝑥 − 2)/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥+4∫1▒𝑑𝑥/(𝑥^2 − 2𝑥 − 5) Rough (𝑥^2−2𝑥−5)^′=2𝑥−2 Solving 𝑰𝟏 I1=1/2 ∫1▒(2𝑥 − 2)/(𝑥^2 − 2𝑥 − 5) . 𝑑𝑥 Let 𝑥^2 − 2𝑥 − 5=𝑡 Diff both sides w.r.t.x 2𝑥−2−0=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(2𝑥 − 2) Thus, our equation becomes ∴ I1=1/2 ∫1▒(2𝑥 − 2)/(𝑥^2 − 2𝑥 − 5) . 𝑑𝑥 Putting value of (𝑥^2−2𝑥−5)=𝑡 and 𝑑𝑥=𝑑𝑡/(2𝑥 − 2) I1=1/2 ∫1▒(2𝑥 − 2)/𝑡 . 𝑑𝑥 I1=1/2 ∫1▒(2𝑥 − 2)/𝑡 . 𝑑𝑡/(2𝑥 − 2) I1=1/2 ∫1▒1/𝑡 . 𝑑𝑡 I1=1/2 log⁡|𝑡|+𝐶1 I1=1/2 log⁡|𝑥^2−2𝑥−5|+𝐶1 Solving 𝑰𝟐 I2=4∫1▒1/(𝑥^2 − 2𝑥 − 5) . 𝑑𝑥 (Using 𝑡=𝑥^2−2𝑥−5) I2=4∫1▒1/(𝑥^2 − 2(𝑥)(1) − 5) . 𝑑𝑥 I2=4∫1▒1/(𝑥^2 − 2(𝑥)(1) + (1)^2 − (1)^2 − 5) . 𝑑𝑥 I2=4∫1▒1/((𝑥 − 1)^2 − (1)^2 − 5) . I2=4∫1▒1/((𝑥 − 1)^2 − 1 − 5) . 𝑑𝑥 I2=4∫1▒1/((𝑥 − 1)^2 − 6) . 𝑑𝑥 I2=4∫1▒1/((𝑥 − 1)^2 −(√6 )^2 ) . 𝑑𝑥 It is of form ∫1▒𝑑𝑥/(𝑥^2 − 𝑎^2 ) =1/2𝑎 log⁡|(𝑥 − 𝑎)/(𝑥 + 𝑎)|+𝐶 ∴ Replacing 𝑥 by (𝑥−1) and a by √6 , we get I2=4/(2√6) log⁡|(𝑥 − 1 − √6)/(𝑥 − 1 + √6)|+𝐶2 I2=2/√6 log⁡|(𝑥 − 1 − √6)/(𝑥 − 1 + √6)|+𝐶2 Putting the values of I1 and I2 in (1) ∫1▒〖(𝑥 + 2)/√(𝑥^2 + 2𝑥 + 3).〗 . 𝑑𝑥 = 𝐼_1+𝐼_2 =1/2 log⁡|𝑥^2−2𝑥−5|+𝐶1+2/√6 log⁡|(𝑥 − 1 − √6)/(𝑥 − 1 + √6)|+𝐶"2 " =𝟏/𝟐 𝒍𝒐𝒈⁡|𝒙^𝟐−𝟐𝒙−𝟓|+𝟐/√𝟔 𝒍𝒐𝒈⁡|(𝒙 − 𝟏 − √𝟔)/(𝒙 − 𝟏 + √𝟔)|+𝑪

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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.