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Ex 7.4
Ex 7.4, 2 Important
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Ex 7.4, 22 You are here
Ex 7.4, 23 Important
Ex 7.4, 24 (MCQ)
Ex 7.4, 25 (MCQ) Important
Last updated at May 29, 2023 by Teachoo
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