Example 16
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 16 Find 𝑥2+ 𝑥 +1 𝑑𝑥 𝑥 + 2 𝑥2+1 By partial fraction 𝑥2+ 𝑥 + 1 𝑥 + 1 𝑥 + 2= 𝐴𝑥 + 2 + 𝐵𝑥 + 𝐶 𝑥2+ 1 Cancelling denominator 𝑥2+ 𝑥+1=𝐴 𝑥2+1+ 𝐵𝑥+𝐶 𝑥+2 Putting x = −𝟐 −22+ −2+1=𝐴 −22+1+0 4−2+1= 5A 35 = A Putting x = 𝟎 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 35 + 2C 1 – 35 = 2C 25 = 2C C = 15 Thus, 𝑥2+ 𝑥 + 1 𝑥 + 1( 𝑥2+ 1) = 35 (𝑥 + 2) + 1 (2𝑥 + 1)5 ( 𝑥2 + 1) Hence, our equation becomes 𝑥2+ 𝑥 + 1 𝑥 + 2 ( 𝑥2 + 1) 𝑑𝑥= 35( 𝑥2 + 1) 𝑑𝑥+ 15 (2𝑥 + 1) 𝑥2 + 1 𝑑𝑥 = 35( 𝑥2 + 1) 𝑑𝑥+ 15 2𝑥 𝑥2 + 1 𝑑𝑥+ 15 1 𝑥2 + 1 𝑑𝑥 Hence 𝑥2+ 𝑥 + 1 𝑥 + 2 ( 𝑥2+ 1) 𝑑𝑥 = 35𝑙𝑜𝑔 𝑥+2+ 15𝑙𝑜𝑔 𝑥2+1+ 15 𝑡𝑎𝑛−1 𝑥+ C Where C = C1+ C2+ C3
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Example 25 (Supplementary NCERT)
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.