Compare the constant term ² 4 0 then we have 5 12 ³

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Compare the constant term: −? + ? − ? + ? = −4 ⇒ ? = 0.Then we have 5?3− 12?2+ 5? − 4(2? + 1)(? − 1)3=32? + 1+1? − 1+0(? − 1)22(? − 1)3The integral can be computed as 5?3− 12?2+ 5? − 4(2? + 1)(? − 1)3?? = 3 ∫12? + 1?? + ∫1? − 1?? − 2 ∫1(? − 1)3??=∫(?−1)−3??=32ln|2? + 1| + ln|? − 1| + (? − 1)−2+ ?.
Page 99of 125Example 33 Compute the integral 6?2− 3? + 7(4? + 1)(?2+ 4)??.Solution: We use the method of partial fractions to decompose the rational function. Since the denominator consists of a quadratic factor ?2+ 4, so we shall try the following decomposition: 6?2− 3? + 7(4? + 1)(?2+ 4)=?4? + 1+?? + ??2+ 4⇒ ?(?2+ 4) + (4? + 1)(?? + ?) = 6?2− 3? + 7. Substitute ? = −14, we get 6516? = 6 (−14)2− 3 (−14) + 7 ⇒ ? = 2.Compare the coefficient of ?2: ? + 4? = 6 ⇒ ? = 1.Compare the constant term: 4? + ? = 7 ⇒ ? = −1.
Page 100of 125Thus we obtain 6?2− 3? + 7(4? + 1)(?2+ 4)=24? + 1+? − 1?2+ 4.The integral can be computed as 6?2− 3? + 7(4? + 1)(?2+ 4)?? = 2 ∫14? + 1?? + ∫? − 1?2+ 4?? … … (∗)The first term can be computed as 14? + 1??=1??+???=1?ln|??+?|+?14ln|4? + 1| + ?.The second term can be computed as ? − 1?2+ 4?? = ∫??2+ 4?? − ∫1?2+ 4??
Page 101of 125=?=?2+4????=2?121??? −141?24+ 1??=12ln|?| −141(?2)2+ 1??=12ln|?2+ 4| −12tan−1?2+ ?.Combing the result, we have 6?2− 3? + 7(4? + 1)(?2+ 4)?? =12ln|4? + 1| +12ln|?2+ 4| −12tan−1?2+ ?.
Page 102of 125Example 34 Compute the integral 6?2+ 9? + 19(? + 1)(?2+ 6? + 13)??10Solution: We use the method of partial fractions to decompose the rational function. Since the factorization of the denominator consists of a quadratic factor (which cannot be factorized further), we shall propose the following decomposition: 6?2+ 9? + 19(? + 1)(?2+ 6? + 13)=?? + 1+?? + ??2+ 6? + 13⇒ ?(?2+ 6? + 13) + (?? + ?)(? + 1) = 6?2+ 9? + 19.Substitute ? = −1, we have 8? = 16 ⇒ ? = 2.Compare the coefficient of ?2:? + ? = 6 ⇒ ? = 4.Compare the constant term: 13? + ? = 19 ⇒ ? = −7.
Page 103of 125Hence, we can compute the integral as 6?2+ 9? + 19(? + 1)(?2+ 6? + 13)??10= 2 ∫1? + 1??10+ ∫4? − 7?2+ 6? + 13??10… … (∗)The first term can be computed as 1? + 1??10= [ln|? + 1|]01= ln 2.The second term can be computed as 4? − 7?2+ 6? + 13??10= ∫4? + 12?2+ 6? + 13??10− ∫19?2+ 6? + 13??10=?=?2+6?+13????=2?+62???2013− 19 ∫1(? + 3)2+ 4??10= 2[ln ?]13201941(? + 3)24+ 1??10= 2[ln 20 − ln 13] −1941(? + 32)2+ 1??10
Page 104of 125= 2[ln 20 − ln 13] − [192tan−1? + 32]01= 2[ln 20 − ln 13] −192tan−12 +192tan−132.Combining the result, we finally get 6?2+ 9? + 19(? + 1)(?2+ 6? + 13)??10= 2 ln 2 + 2[ln 20 − ln 13] −192tan−12 +192tan−132= ln (1600169) −192tan−12 +192tan−132.
Page 105of 125Example 35 (Integral of improper rational function) Compute the integral 3?4+ 3?3− 5?

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