2x 5 3 X 2

Ex seven.4, 22 - Chapter 7 Class 12 Integrals (Term 2)

Last updated at Aug. xx, 2021 by Teachoo

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Ex 7.4, 22 Integrate the function (𝑥 + 3)/(𝑥^2 − 2𝑥 − v) ∫1▒(𝑥 + 3)/(𝑥^2 − 2𝑥 − five) 𝑑𝑥 =i/two ∫i▒(two𝑥 + vi)/(𝑥^2 − 2𝑥 − five) 𝑑𝑥 =ane/2 ∫i▒(2𝑥 − two + ii + 6 )/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥 =1/two ∫1▒(ii𝑥 − 2)/(𝑥^2 − 2𝑥 − 5) 𝑑𝑥+8/2 ∫1▒𝑑𝑥/(𝑥^2 − 2𝑥 − five) =1/two ∫1▒(ii𝑥 − two)/(𝑥^ii − 2𝑥 − 5) 𝑑𝑥+4∫one▒𝑑𝑥/(𝑥^2 − 2𝑥 − 5) Rough (𝑥^2−2𝑥−v)^′=2𝑥−2 Solving 𝑰𝟏 I1=1/2 ∫1▒(two𝑥 − 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/(ii𝑥 − 2) Thus, our equation becomes ∴ I1=1/2 ∫1▒(2𝑥 − ii)/(𝑥^ii − two𝑥 − 5) . 𝑑𝑥 Putting value of (𝑥^ii−2𝑥−5)=𝑡 and 𝑑𝑥=𝑑𝑡/(2𝑥 − 2) I1=1/2 ∫1▒(2𝑥 − 2)/𝑡 . 𝑑𝑥 I1=one/2 ∫1▒(2𝑥 − ii)/𝑡 . 𝑑𝑡/(2𝑥 − 2) I1=1/2 ∫1▒i/𝑡 . 𝑑𝑡 I1=1/ii log⁡|𝑡|+𝐶one I1=1/two log⁡|𝑥^2−2𝑥−five|+𝐶1 Solving 𝑰𝟐 I2=iv∫1▒1/(𝑥^2 − two𝑥 − v) . 𝑑𝑥 (Using 𝑡=𝑥^2−2𝑥−5) I2=4∫1▒1/(𝑥^2 − 2(𝑥)(1) − 5) . 𝑑𝑥 I2=4∫i▒1/(𝑥^two − 2(𝑥)(i) + (1)^2 − (1)^2 − 5) . 𝑑𝑥 I2=4∫i▒i/((𝑥 − 1)^2 − (1)^ii − v) . I2=4∫1▒1/((𝑥 − i)^2 − 1 − v) . 𝑑𝑥 I2=4∫1▒1/((𝑥 − i)^2 − 6) . 𝑑𝑥 I2=4∫1▒1/((𝑥 − i)^ii −(√six )^2 ) . 𝑑𝑥 Information technology is of form ∫ane▒𝑑𝑥/(𝑥^ii − 𝑎^two ) =1/2𝑎 log⁡|(𝑥 − 𝑎)/(𝑥 + 𝑎)|+𝐶 ∴ Replacing 𝑥 by (𝑥−1) and a by √6 , we go I2=4/(2√half dozen) log⁡|(𝑥 − i − √6)/(𝑥 − 1 + √6)|+𝐶2 I2=two/√6 log⁡|(𝑥 − 1 − √6)/(𝑥 − i + √half-dozen)|+𝐶two Putting the values of I1 and I2 in (i) ∫1▒〖(𝑥 + 2)/√(𝑥^two + 2𝑥 + 3).〗 . 𝑑𝑥 = 𝐼_1+𝐼_2 =1/2 log⁡|𝑥^2−2𝑥−5|+𝐶1+2/√6 log⁡|(𝑥 − i − √6)/(𝑥 − 1 + √6)|+𝐶"2 " =𝟏/𝟐 𝒍𝒐𝒈⁡|𝒙^𝟐−𝟐𝒙−𝟓|+𝟐/√𝟔 𝒍𝒐𝒈⁡|(𝒙 − 𝟏 − √𝟔)/(𝒙 − 𝟏 + √𝟔)|+𝑪

2x 5 3 X 2,

Source: https://www.teachoo.com/5019/719/Ex-7.4--22---Integrate-x---3---x2---2x---5/category/Ex-7.4/

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