Methods in calculus

FM Methods in Calculus: inverse trig functions KUS objectives BAT integrate using partial fractions Starter: express in partial fractions 1 𝑥 − 1 𝑥+1 1 𝑥 𝑥+1 2𝑥−1 𝑥 2 −4 5 4(𝑥+2) + 3 4(𝑥−2) 3𝑥−2 2 𝑥 2 +3𝑥+1 5 𝑥+1 − 7 2𝑥+1

Don’t mix up 1 𝑎 2 − 𝑥 2 𝑑𝑥 and 1 𝑎 2 + 𝑥 2 𝑑𝑥 WB E1 a) show that 1 𝑎 2 − 𝑥 2 𝑑𝑥= 1 2𝑎 ln 𝑎+𝑥 𝑎−𝑥 +𝐶 where a is a constant 1 𝑎 2 − 𝑥 2 𝑑𝑥= 1 (𝑎+𝑥)(𝑎−𝑥) 𝑑𝑥 1 (𝑎+𝑥)(𝑎−𝑥) = 1 2𝑎 1 (𝑎+𝑥) + 1 (𝑎−𝑥) = 1 2a 1 (𝑎+𝑥) 𝑑𝑥 + 1 2𝑎 1 (𝑎−𝑥) 𝑑𝑥 = 1 2𝑎 ln 𝑎+𝑥 + ln 𝑎−𝑥 +𝐶 = 1 2𝑎 ln 𝑎+𝑥 𝑎−𝑥 +𝐶 Don’t mix up 1 𝑎 2 − 𝑥 2 𝑑𝑥 and 1 𝑎 2 + 𝑥 2 𝑑𝑥

5 𝑥 2 +𝑥−10 (𝑥+1) 𝑥 2 −3 = 𝑨 𝑥+1 + 𝑩𝑥+𝐶 𝑥 2 −3 Notes partial fractions and quadratic factors When a partial fraction includes a quadratic factor of the form 𝑥 2 +𝑐 we need to put a linear numerator of the form 𝐴𝑥+𝐵 on the partial fraction 5 𝑥 2 +𝑥−10 (𝑥+1) 𝑥 2 −3 = 𝑨 𝑥+1 + 𝑩𝑥+𝐶 𝑥 2 −3 WB E2 5 𝑥 2 +𝑥−10=𝑨 𝑥 2 −3 + 𝑩𝑥+𝑪 𝑥+1 Let x=−1 then 5+(−1)−10=𝑨 1−3 gives 𝑨=𝟑 Look at coefficients of 𝑥 2 5 𝑥 2 =𝑨 𝑥 2 +𝑩 𝑥 2 gives 𝑩=𝟐 𝑥=𝑩𝑥+𝑪𝑥 gives 𝑪=−𝟏 Look at coefficients of 𝑥 5 𝑥 2 +𝑥−10 (𝑥+1) 𝑥 2 −3 = 𝟑 𝑥+1 + 𝟐𝑥−𝟏 𝑥 2 −3

= 1 18 ln 𝑥 2 𝑥 2 +9 − 1 3 arctan 𝑥 3 +𝐶 So 𝐴= 1 18 𝐵= 1 3 WB E3 Show that 1+𝑥 𝑥 3 +9𝑥 𝑑𝑥=𝐴 ln 𝑥 2 𝑥 2 +9 +𝐵𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 3 +𝐶 where A and B are constants to be found 1+𝑥 𝑥 3 +9𝑥 𝑑𝑥= 1+𝑥 𝑥(𝑥 2 +9) 𝑑𝑥 1+𝑥 𝑥(𝑥 2 +9) = 1 9 1 𝑥 + 𝑥−9 𝑥 2 +9 = 1 9 1 𝑥 𝑑𝑥 + 1 9 𝑥 𝑥 2 +9 𝑑𝑥 − 1 9 9 𝑥 2 +9 𝑑𝑥 1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶 = 1 9 ln 𝑥 + 1 9 1 2 ln 𝑥 2 +9 + 1 3 arctan 𝑥 3 +C = 1 18 2 ln 𝑥 + 1 18 ln 𝑥 2 +9 + 1 3 arctan 𝑥 3 +𝐶 = 1 18 ln 𝑥 2 𝑥 2 +9 − 1 3 arctan 𝑥 3 +𝐶 So 𝐴= 1 18 𝐵= 1 3

Look at coefficients of 𝑥 4 WB E4 a) express 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 as partial fractions b) Hence find 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 𝑑𝑥 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 = 𝑥 4 +𝑥 𝑥 2 +2 𝑥 2 +3 =A+ 𝐵𝑥+𝐶 𝑥 2 +2 + 𝐷𝑥+𝐸 𝑥 2 +3 𝑥 4 +𝑥=𝐴 𝑥 2 +2 𝑥 2 +3 + 𝐵𝑥+𝐶 𝑥 2 +3 +(𝐷𝑥+𝐸) 𝑥 2 +2 Look at coefficients of 𝑥 4 𝑥 4 =𝑨 𝑥 4 gives 𝑨=𝟏 Look at coefficients of 𝑥 3 0=𝐵 𝑥 3 +𝐷 𝑥 3 gives 𝑩+𝑫=𝟎 Look at coefficients of 𝑥 𝑥=3𝐵𝑥+2𝐷𝑥 gives 3𝑩+𝟐𝑫=𝟏 Solve these simultaneous equations gives 𝑩=𝟏 𝑫=−𝟏 Look at coefficients of 𝑥 2 0=5𝐴 𝑥 2 +𝐶 𝑥 2 +𝐸 𝑥 2 gives C+𝐸=−𝟓 Look at constant terms 0=6𝐴+3𝐶+2𝐸 gives 3C+2𝐸=−𝟔 Solve these simultaneous equations gives 𝑪=𝟒 𝑬=−𝟗

WB E5 a) express 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 as partial fractions b) Hence find 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 𝑑𝑥 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 = 𝑥 4 +𝑥 𝑥 2 +2 𝑥 2 +3 =1+ 𝑥+4 𝑥 2 +2 + −𝑥−9 𝑥 2 +3 𝑏) 𝑥 4 +𝑥 𝑥 4 +5 𝑥 2 +6 𝑑𝑥 = 1 𝑑𝑥+ 𝑥 𝑥 2 +2 𝑑𝑥+ 4 𝑥 2 +2 𝑑𝑥− 𝑥 𝑥 2 +3 𝑑𝑥 − 9 𝑥 2 +3 𝑑𝑥 = 𝑥+ 1 2 ln 𝑥 2 +2 + 4 2 arctan 𝑥 2 − 1 2 ln 𝑥 2 +3 − 9 3 arctan 𝑥 3 +𝐶 = 𝑥+ 1 2 ln 𝑥 2 +2 𝑥 2 +3 + 4 2 arctan 𝑥 2 − 9 3 arctan 𝑥 3 +𝐶 NOW DO EX 3E

One thing to improve is – KUS objectives BAT integrate using partial fractions self-assess One thing learned is – One thing to improve is –

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