x 3 – 3c . x + 3 x2 + 0x + 4 x2 + 3x Correct division or Correct 3x + 4 – similar = u Integration 3x 9 – – evaluated = r. + 13 = = x2 3x + 13 ln(x + 3) + c – 2 3d I = Let u = x + 2 so x = u 2 and x 2 = u 4 – – – du = dx A I = equ Uses substitution to = obtain equ A = 8 – = 4 Evaluates integral correctly
3c x – 3 . x + 3 x2 + 0x + 4 x2 + 3x – 3x + 4 – 3x – 9 + 13 = = x2 – 3x + 13 ln(x + 3) + c 2 Correct division or similar = u Correct Integration evaluated = r. 3d I = Let u = x + 2 so x = u – 2 and x – 2 = u – 4 du = dx I = equ A = = – 8 = 4 Uses substitution to obtain equ A Evaluates integral correctly
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