Example 13 - Chapter 7 Class 12 Integrals
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 13 Find ∫1▒(3𝑥 −2)/((𝑥 + 1)^2 (𝑥 + 3) ) 𝑑𝑥 We can write Integral as (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=𝐴/(𝑥 + 1) + 𝐵/(𝑥 + 1)^2 + 𝐶/((𝑥 + 3) ) (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=(𝐴(𝑥 + 1)(𝑥 + 3) + 𝐵(𝑥 + 3) + 𝐶(𝑥 + 1)^2)/((𝑥 + 1)^2 (𝑥 + 3) ) Cancelling denominator 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 Putting x = −1 3(−1) −2=𝐴(−1+1)(−1+3)+𝐵(−1+3)+𝐶(−1+1)^2 −3−2=𝐴×0+𝐵×2+𝐶×(0)^2 −5=𝐵×2 𝐵=(− 5)/2 Putting x = − 3 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 3(−3)−2=𝐴(−3+1)(−3+3)+𝐵(−3+3)+𝐶(−3+1)^2 −9−2=𝐴×0+𝐵×0+𝐶×(−2)^2 −11=0+0+𝐶(4) −11=4𝐶 (−11)/4 =𝐶 𝐶 =(−11)/4 Putting x = 0 3𝑥 −2=𝐴(𝑥+1)(𝑥+3)+𝐵(𝑥+3)+𝐶(𝑥+1)^2 3(0) − 2 = A(1) (3) + B(3) + C 〖"(1)" 〗^2 −2 = 3A + 3B + C Putting value of B & C −2 = 3A + 3((−5)/2) + ((−11)/4) −2 = 3A − 15/2−11/4 −2 = 3A + (−30 − 11)/4 −8 = 12A − 41 41 − 8 = 12A 33/12 = A 11/4 = A Hence, we can write (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=𝐴/(𝑥 + 1) + 𝐵/(𝑥 + 1)^2 + 𝐶/((𝑥 + 3) ) (3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) )=11/4(𝑥 + 1) − 5/〖2(𝑥 + 1)〗^2 − 11/4(𝑥 + 3) Therefore ∫1▒(3𝑥 − 2)/((𝑥 + 1)^2 (𝑥 + 3) ) 𝑑𝑥 =∫1▒11/4(𝑥 + 1) 𝑑𝑥−∫1▒5/〖2(𝑥 + 1)〗^2 𝑑𝑥−∫1▒11/4(𝑥 + 3) 𝑑𝑥 =11/4 log|𝑥+1|−5/2×((−1))/((𝑥 + 1) ) − 11/4 log|𝑥+3|+𝐶 =11/4 (log|𝑥+1|−log|𝑥+3| )+5/(2 (𝑥 + 1) )+𝐶 =𝟏𝟏/𝟒 𝒍𝒐𝒈|(𝒙 + 𝟏)/(𝒙 + 𝟑)| + 𝟓/(𝟐 (𝒙 + 𝟏) )+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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo