51 2 1 3 1 ln 2 2 = 2 3 1 2  c ln x dx 2 2 1 dx 2 ln  1 2 x a  a  x c  x a Hence 2 dx Note: since x2 3x + 2 = (x 2) (x 1), we can resolve – – – 3 2  x x into partial 2 fraction and then integrate to get the same answer. Try this method yourself. Example 16 dx Find 2 5 2  x x 2 Solution 5 Let 2x2 + 5x + 1 = 2(x2 + = 2[(x + a)2 a2 + 1] – = 2[x2 + 2ax + a2 a2 + 1] – 5 Comparing the coefficient of x on both 4a = 5 or a = . 4 2 2 2  x  2 4 4 2  9 x 4 = 16 2
51 = 1 2 ln 3 2 3 2 1 2 1 2  c Hence 2 dx  ln dx 2 x 2 x 1  c 2  1 2 a ln x a x  a Note: since x2 – 3x + 2 = (x – 2) (x – 1), we can resolve dx x 2 3 x  2 into partial fraction and then integrate to get the same answer. Try this method yourself. Example 16 Find dx 2 x 2 5 x  2 Solution 5 Let 2x2 + 5x + 1 = 2(x2 + = 2[(x + a)2 – a2 + 1] = 2[x2 + 2ax + a2 – a2 + 1] Comparing the coefficient of x on both 4a = 5 or a = 5 4 . 2 2 x  2 2  4 4 = 2 x  2 4 9 16
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