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∫(4x+1)2 dx

The answer can be [(4x+1)2/12]+C .

But if I expand the equation to integrate it,I get different answer,which is 16x3/3 + 4x2 + x + C

Is both answer acceptable in exam?

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Please take note that your first answer is wrong, it should be [(4x + 1)2/3] + C 

∫(4x+1)2 dx can be solved in two methods:

First method

Let u = 4x + 1

∫(4x+1)2 dx = ∫u2 dx
                 =  [u (2+1)/(2+1)]  + C
                 =  [u2/3] + C
                 = [(4x + 1)2/3] + C  (replace back u into equation)

2nd method, expand the equation (4x + 1)2

∫(4x+1)2 dx = ∫ 16x2 + 8x + 1  dx
                =  16x3/3 + 4x2 + x + C 

1st method answer =>  [(4x + 1)2/3] + C
2nd method answer => 16x3/3 + 4x2 + x + C

C is a unknown constant in both equation and the value may not be necessary the same.

so, if you want to check the equality for both answer, you can further expand the first answer  [(4x + 1)2/3] + C

[(4x + 1)2/3] + C = [(4x+1)(4x + 1)2/3] + C
                        = [(4x+1)(16x2 + 8x + 1)/3] + C
                        = [ (64x3 + 16x2 + 32x2 + 8x + 4x + 1)/3 ] + C
                        = [ 64x3 + 48x2 + 12x + 1)/3 ] + C
                        =  64x3/3 + 16x2 + 4x + 1/3 + C
                        =  64x3/3 + 16x2 + 4x + D  ( C is unknown constant, so we can use D = [C + 1/3], D is also an unknown constant

So, first answer =  [(4x + 1)2/3] + C  = 64x3/3 + 16x2 + 4x + D

Is 64x3/3 + 16x2 + 4x + D (expansion of 1st answer)  equals to 2nd answer 16x3/3 + 4x2 + x + C ?

To find out, try times 4 with the second answer 16x3/3 + 4x2 + x + C :
4 x (
16x3/3 + 4x2 + x + C)  =  64x3/3 + 16x2 + 4x + 4C
                                       =  
64x3/3 + 16x2 + 4x + D  (D=4C)

So, first answer  [(4x + 1)2/3] + C  = 64x3/3 + 16x2 + 4x + D
Second answer (after times 4) = 
64x3/3 + 16x2 + 4x + D

Both answers are equal. Thus to answer your question, both methods (without expanding and expanding) are acceptable.

用户: (46,890 分) 15 50 59
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