Solution dx dx = 4 25 4  x 25 2 5 dx 1 2 2 25 = 2 1 = 2 x  25 c tan 5 Example 15 2 Find x dx 3 2  x Solution x2 3x + 2 = x2 3x + a2 2 – – –a2 – = (x a)2 a2 + 2 – – = x2 2ax + a2 a2 + 2 – – 3 Equating coefficient of x on both sides 2a = 3 or a = 2 2 2 2  2 – 2 2 1 2 4 dx x 2 2 Then 2 2 2 dx =
Solution dx 25 x 2  4 = dx 4 25 1 = 25 dx 2 2 5 2 = 1 25 tan 2 x 5  c Example 15 Find x 2 dx 3 x  2 Solution x2 – 3x + 2 = x2 – 3x + a2 –a2 – 2 = (x – a)2 – a2 + 2 = x2 – 2ax + a2 – a2 + 2 Equating coefficient of x on both sides 2a = 3 or a = 3 2 2 2 2 2 –  2 2 2 1 4 Then 2 dx = x dx 2 2 2 2
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