Integration by partial fraction - Type 3

Chapter 7 Class 12 Integrals
Concept wise

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### Transcript

Ex 7.5, 11 Integrate the function 5𝑥/((𝑥 + 1) (𝑥2− 4) ) We can write the integrand as 5𝑥/((𝑥 + 1) (𝑥2− 4) ) = 5𝑥/((𝑥 + 1) (𝑥 − 2) (𝑥 + 2) ) 5𝑥/((𝑥 + 1) (𝑥2− 4) ) = 𝐴/((𝑥 + 1) ) + 𝐵/((𝑥 − 2) ) + 𝐶/((𝑥 + 2) ) 5𝑥/((𝑥 + 1) (𝑥2− 4) ) = (𝐴(𝑥 − 2)(𝑥 + 2) + 𝐵(𝑥 + 1)(𝑥 + 2) + 𝐶(𝑥 +1)(𝑥 − 2))/((𝑥 + 1) (𝑥 − 2) (𝑥 + 2) ) Cancelling denominator 5𝑥 = 𝐴(𝑥−2)(𝑥+2)+𝐵(𝑥+1)(𝑥+2)+𝐶(𝑥+1)(𝑥−2) …(1) Putting x = −1 in (1) 5𝑥 = 𝐴(𝑥−2)(𝑥+2)+𝐵(𝑥+1)(𝑥+2)+𝐶(𝑥+1)(𝑥−2) 5( −1) = 𝐴(−1−2)(−1+2)+𝐵(−1+1)(−1+2)+𝐶(−1+1)(−1−2) −5 = 𝐴(−3)(1)+𝐵×0+𝐶×0 −5 = −3𝐴 𝐴 = (−5)/(−3) = 5/3 Putting x = 2 in (1) 5𝑥 = 𝐴(𝑥−2)(𝑥+2)+𝐵(𝑥+1)(𝑥+2)+𝐶(𝑥+1)(𝑥−2) 5"(2) = " 𝐴(2−2)(2+2)+𝐵(2+1)(2+2)+𝐶(2+1)(2−2) 10 = 𝐴×0+𝐵(3)(4)+𝐶×0 10 = 12𝐵 𝐵 = 10/12=5/6 Putting x = −2 in (1) 5𝑥 = 𝐴(𝑥−2)(𝑥+2)+𝐵(𝑥+1)(𝑥+2)+𝐶(𝑥+1)(𝑥−2) 5"("−"2) = " 𝐴(−2−2)(−2+2)+𝐵(−2+1)(−2+2)+𝐶(−2+1)(−2−2) −10 = 𝐴×0+𝐵×0+𝐶(−1)(−4) −10 = 4𝐶 𝐶 = (−10)/4 𝐶 = (−5)/2 Therefore ∫1▒5𝑥/((𝑥 + 1) (𝑥2− 4) )=∫1▒(𝐴/(𝑥 + 1)+𝐵/(𝑥 − 2)+𝐶/(𝑥 + 2)) 𝑑𝑥 =5/3 ∫1▒𝑑𝑥/(𝑥 + 1) 𝑑𝑥+ 5/6 ∫1▒𝑑𝑥/(𝑥 − 2) 𝑑𝑥−5/2 ∫1▒𝑑𝑥/((𝑥 + 2) ) =𝟓/𝟑 〖𝒍𝒐𝒈 〗⁡|𝒙+𝟏|− 𝟓/𝟐 〖𝐥𝐨𝐠 〗⁡|𝒙+𝟐|+𝟓/𝟔 〖𝐥𝐨𝐠 〗⁡|𝒙−𝟐|+𝑪